calculo.libro week.1 5
TRANSCRIPT
MOOCULUSmassive open online calculus
C A L C U L U S
T H I S D O C U M E N T W A S T Y P E S E T O N F E B R U A R Y 4 , 2 0 1 3 .
2
Copyright c� 2013 Jim Fowler and Bart Snapp
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Street, 5th Floor, San Francisco, California, 94105, USA. If you distribute this work or a derivative, include the history of the
document.
This text is based on David Guichard’s open-source calculus text which in turn is a modification and expansion of notes written by
Neal Koblitz at the University of Washington. David Guichard’s text is available at http://www.whitman.edu/mathematics/
calculus/ under a Creative Commons license.
The book includes some exercises and examples from Elementary Calculus: An Approach Using Infinitesimals, by H. Jerome Keisler,
available at http://www.math.wisc.edu/~keisler/calc.html under a Creative Commons license. In addition, the chapter
on differential equations and the section on numerical integration are largely derived from the corresponding portions of Keisler’s
book. Albert Schueller, Barry Balof, and Mike Wills have contributed additional material.
This book is typeset in the Kerkis font, Kerkis c� Department of Mathematics, University of the Aegean.
We will be glad to receive corrections and suggestions for improvement at [email protected] or [email protected].
Contents
1 Limits 6
2 Basics of Derivatives 35
3 Curve Sketching 53
4 The Product Rule and Quotient Rule 71
5 The Chain Rule 79
Answers to Exercises 96
Index 104
List of Main Theorems
1.3.1 Theorem (Limit Laws) . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.3.5 Theorem (Squeeze Theorem) . . . . . . . . . . . . . . . . . . . . . . 19
1.6.3 Theorem (Intermediate Value Theorem) . . . . . . . . . . . . . . . . 31
2.1.3 Theorem (Differentiability implies Continuity) . . . . . . . . . . . . 38
2.2.1 Theorem (The Constant Rule) . . . . . . . . . . . . . . . . . . . . . 44
2.2.2 Theorem (The Power Rule) . . . . . . . . . . . . . . . . . . . . . . . 45
2.2.6 Theorem (The Sum Rule) . . . . . . . . . . . . . . . . . . . . . . . . 47
2.2.9 Theorem (The Derivative of ex) . . . . . . . . . . . . . . . . . . . . . 49
3.1.1 Theorem (Fermat’s Theorem) . . . . . . . . . . . . . . . . . . . . . 54
3.2.1 Theorem (First Derivative Test) . . . . . . . . . . . . . . . . . . . . 57
3.3.1 Theorem (Test for Concavity) . . . . . . . . . . . . . . . . . . . . . 61
3.4.1 Theorem (Second Derivative Test) . . . . . . . . . . . . . . . . . . . 64
4.1.1 Theorem (The Product Rule) . . . . . . . . . . . . . . . . . . . . . . 72
4.2.1 Theorem (The Quotient Rule) . . . . . . . . . . . . . . . . . . . . . 75
5.1.1 Theorem (Chain Rule) . . . . . . . . . . . . . . . . . . . . . . . . . 79
5.2.2 Theorem (The Derivative of the Natural Logrithm) . . . . . . . . . . 88
5.2.3 Theorem (Inverse Function Theorem) . . . . . . . . . . . . . . . . . 89
How to Read Mathematics
Reading mathematics is not the same as reading a novel. To read mathematics you
need:
(a) A pen.
(b) Plenty of blank paper.
(c) A willingness to write things down.
As you read mathematics, you must work along side of the text itself. You must
write down each expression, sketch each graph, and think about what you are
doing. You should work examples and fill-in the details. This is not an easy task, it
is in fact hard work. However, mathematics is not a passive endeavor. You, the
reader, must become a doer of mathematics.
1 Limits
1.1 The Basic Ideas of Limits
Consider the function:
f (x) =x2 − 3x + 2
x − 2
While f (x) is undefined at x = 2, we can still plot f (x) at other values, see Figure 1.1.
Examining Table 1.1, we see that as x approaches 2, f (x) approaches 1. We write
this:
As x → 2, f (x)→ 1 or limx→2
f (x) = 1.
Intuitively, limx→a
f (x) = L when the value of f (x) can be made arbitrarily close to L
by making x sufficiently close, but not equal to, a. This leads us to the formal
definition of a limit.
−2 −1 1 2 3 4
−3
−2
−1
1
2
3
x
y
Figure 1.1: A plot of f (x) =x2 − 3x + 2
x − 2.
x f (x)
1.7 0.71.9 0.91.99 0.991.999 0.9992 undefined
x f (x)
2 undefined2.001 1.0012.01 1.012.1 1.12.3 1.3
Table 1.1: Values of f (x) =x2 − 3x + 2
x − 2.
Equivalently, limx→a
f (x) = L, if for every ε > 0 there is
a δ > 0 so that whenever x , a and a−δ < x < a+δ,we have L − ε < f (x) < L + ε.
Definition The limit of f (x) as x goes to a is L,
limx→a
f (x) = L,
if for every ε > 0 there is a δ > 0 so that whenever
0 < |x − a| < δ, we have |f (x) − L | < ε.
If no such value of L can be found, then we say that limx→a
f (x) does not exist.
In Figure 1.2, we see a geometric interpretation of this definition.
calculus 7
a − δ a a + δ
L − ε
L
L + ε
x
y Figure 1.2: A geometric interpretation of the (ε, δ)-criterion for limits. If 0 < |x − a| < δ, then we havethat a − δ < x < a + δ. In our diagram, we see thatfor all such x we are sure to have L − ε < f (x) < L + ε,and hence |f (x) − L | < ε.
Limits need not exist, let’s examine two cases of this.
Example 1.1.1 Let f (x) = ⌊x⌋. Explain why the limit
limx→2
f (x)
does not exist.
−2 −1 1 2 3 4
−2
−1
1
2
3
x
y
Figure 1.3: A plot of f (x) = ⌊x⌋. Note, no matterwhich δ > 0 is chosen, we can only at best boundf (x) in the interval [1, 2].
Solution The function ⌊x⌋ is the function that returns the greatest integer less
than or equal to x. Since f (x) is defined for all real numbers, one might be
tempted to think that the limit above is simply f (2) = 2. However, this is not the
case. If x < 2, then f (x) = 1. Hence if ε = .5, we can always find a value for x
(just to the left of 2) such that
0 < |x − 2| < δ, where ε < |f (x) − 2|.
On the other hand, limx→2
f (x) , 1, as in this case if ε = .5, we can always find a
value for x (just to the right of 2) such that
0 < |x − 2| < δ, where ε < |f (x) − 1|.
8
We’ve illustrated this in Figure 1.3. Moreover, no matter what value one chooses
for limx→2
f (x), we will always have a similar issue.
With the example of f (x) = ⌊x⌋, we see that takinglimits is truly different from evaluating functions.
Limits may not exist even if the formula for the function looks innocent.
Example 1.1.2 Let f (x) = sin
(1
x
)
. Explain why the limit
limx→0
f (x)
does not exist.
Solution In this case f (x) oscillates “wildly” as x approaches 0, see Figure 1.4.
In fact, one can show that for any given δ, There is a value for x in the interval
0 − δ < x < 0 + δ
such that f (x) is any value in the interval [−1, 1]. Hence the limit does not exist.
−0.2 −0.1 0.1 0.2x
y
Figure 1.4: A plot of f (x) = sin
(1
x
)
.
Sometimes the limit of a function exists from one side or the other (or both)
even though the limit does not exist. Since it is useful to be able to talk about this
situation, we introduce the concept of a one-sided limit:
Definition We say that the limit of f (x) as x goes to a from the left is L,
limx→a−
f (x) = L
if for every ε > 0 there is a δ > 0 so that whenever x < a and
a − δ < x we have |f (x) − L | < ε.
We say that the limit of f (x) as x goes to a from the right is L,
limx→a+
f (x) = L
calculus 9
if for every ε > 0 there is a δ > 0 so that whenever x > a and
x < a + δ we have |f (x) − L | < ε.Limits from the left, or from the right, are collectivelycalled one-sided limits.
Example 1.1.3 Let f (x) = ⌊x⌋. Discuss
limx→2−
f (x), limx→2+
f (x), and limx→2
f (x).
Solution From the plot of f (x), see Figure 1.3, we see that
limx→2−
f (x) = 1, and limx→2+
f (x) = 2.
Since these limits are different, limx→2
f (x) does not exist.
10
Exercises for Section 1.1
(1) Evaluate the expressions by reference to the plot in Figure 1.5.
-4 -2 2 4 6
-2
2
4
6
8
10
x
y
Figure 1.5: A plot of f (x), a piecewise defined func-tion.
(a) limx→4
f (x)
(b) limx→−3
f (x)
(c) limx→0
f (x)
(d) limx→0−
f (x)
(e) limx→0+
f (x)
(f) f (−2)
(g) limx→2−
f (x)
(h) limx→−2−
f (x)
(i) limx→0
f (x + 1)
(j) f (0)
(k) limx→1−
f (x − 4)
(l) limx→0+
f (x − 2)
➠
(2) Use a table and a calculator to estimate limx→0
sin(x)
x. ➠
(3) Use a table and a calculator to estimate limx→0
sin(2x)
x. ➠
(4) Use a table and a calculator to estimate limx→0
x
sin(
x3
) . ➠
(5) Use a table and a calculator to estimate limx→0
tan(3x)
tan(5x).
➠
(6) Use a table and a calculator to estimate limx→0
2x − 1
x. ➠
(7) Use a table and a calculator to estimate limx→0
(1 + x)1/x . ➠
(8) Sketch a plot of f (x) =x
|x |and explain why lim
x→0
x
|x |does not exist. ➠
(9) Let f (x) = sin
(π
x
)
. Construct three tables of the following form
x f (x)
0.d
0.0d
0.00d
0.000d
calculus 11
where d = 1, 3, 7. What do you notice? How do you reconcile the entries in your
tables with the value of limx→0
f (x)? ➠
(10) In the theory of special relativity, a moving clock ticks slower than a stationary
observer’s clock. If the stationary observer records that ts seconds have passed,
then the clock moving at velocity v has recorded that
tv = ts
√
1 − v2/c2
seconds have passed, where c is the speed of light. What happens as v → c
from below? ➠
12
1.2 Limits by the Definition
Now we are going to get our hands dirty, and really use the definition of a limit. Recall, limx→a
f (x) = L, if for every ε > 0 there is a
δ > 0 so that whenever 0 < |x − a| < δ, we have|f (x) − L | < ε.
2 − δ 2 2 + δ
4 − ε
4
4 + ε
x
y
Figure 1.6: The (ε, δ)-criterion for limx→2
x2= 4. Here
δ = min
(ε
5, 1
)
.
Example 1.2.1 Show that limx→2
x2= 4.
Solution We want to show that for any given ε > 0, we can find a δ > 0 such
that
|x2 − 4| < ε
whenever 0 < |x − 2| < δ. Start by factoring the left-hand side of the inequality
above
|x + 2||x − 2| < ε.
Since we are going to assume that 0 < |x − 2| < δ, we will focus on the factor
|x + 2|. Since x is assumed to be close to 2, suppose that x ∈ [1, 3]. In this case
|x + 2| ≤ 3 + 2 = 5,
and so we want
5 · |x − 2| < ε
|x − 2| <ε
5
Recall, we assumed that x ∈ [1, 3], which is equivalent to |x − 2| < 1. Hence we
must set δ = min
(ε
5, 1
)
.
When dealing with limits of polynomials, the general strategy is always the same.
Let p(x) be a polynomial. If showing
limx→a
p(x) = L,
one must first factor out |x − a| from |p(x) − L |. Next bound x ∈ [a − 1, a + 1] and
estimate the largest possible value of
∣∣∣∣∣
p(x) − L
x − a
∣∣∣∣∣
calculus 13
for x ∈ [a − 1, a + 1]. Call this estimation M . Finally, one must set δ = min
(ε
M, 1
)
.
As you work with limits, you find that you need to do the same procedures again
and again. The next theorems will expedite this process.
Theorem 1.2.2 (Limit Product Law) Suppose limx→a
f (x) = L and limx→a
g(x) = M .
Then
limx→a
f (x)g(x) = LM.
We will use this same trick again of “adding 0” in theproof of Theorem 4.1.1.
This is all straightforward except perhaps for the“≤”. This follows from the Triangle Inequality. TheTriangle Inequality states: If a and b are any realnumbers then |a + b| ≤ |a| + |b|.
Proof Given any ε we need to find a δ such that
0 < |x − a| < δ
implies
|f (x)g(x) − LM | < ε.
Here we use an algebraic trick, add 0 = −f (x)M + f (x)M :
|f (x)g(x) − LM | = |f (x)g(x)−f (x)M + f (x)M − LM |
= |f (x)(g(x) −M) + (f (x) − L)M |
≤ |f (x)(g(x) −M)| + |(f (x) − L)M |
= |f (x)||g(x) −M | + |f (x) − L ||M |.
Since limx→a
f (x) = L, there is a value δ1 so that 0 < |x −a| < δ1 implies |f (x)− L | <|ε/(2M)|. This means that 0 < |x − a| < δ1 implies |f (x) − L ||M | < ε/2.
|f (x)g(x) − LM | ≤ |f (x)||g(x) −M | + |f (x) − L ||M |︸ ︷︷ ︸
<ε
2
.
If we can make |f (x)||g(x)−M | < ε/2, then we’ll be done. We can make |g(x)−M |smaller than any fixed number by making x close enough to a. Unfortunately,
ε/(2f (x)) is not a fixed number since x is a variable.
14
Here we need another trick. We can find a δ2 so that |x − a| < δ2 implies that
|f (x) − L | < 1, meaning that L − 1 < f (x) < L + 1. This means that |f (x)| < N,
where N is either |L −1| or |L +1|, depending on whether L is negative or positive.
The important point is that N doesn’t depend on x. Finally, we know that there
is a δ3 so that 0 < |x − a| < δ3 implies |g(x) −M | < ε/(2N). Now we’re ready to
put everything together. Let δ be the smallest of δ1, δ2, and δ3. Then |x − a| < δ
implies that
|f (x)g(x) − LM | ≤ |f (x)|︸︷︷︸
<N
|g(x) −M |︸ ︷︷ ︸
<ε
2N
+ |f (x) − L ||M |︸ ︷︷ ︸
<ε
2
.
so
|f (x)g(x) − LM | ≤ |f (x)||g(x) −M | + |f (x) − L ||M |
< Nε
2N+
∣∣∣∣∣
ε
2M
∣∣∣∣∣|M |
=ε
2+
ε
2= ε.
This is just what we needed, so by the definition of a limit, limx→a
f (x)g(x) = LM.
Another useful way to put functions together is composition. If f (x) and g(x)
are functions, we can form two functions by composition: f (g(x)) and g(f (x)). For
example, if f (x) =√
x and g(x) = x2+ 5, then f (g(x)) =
√x2 + 5 and g(f (x)) =
(√
x)2+ 5 = x + 5. This brings us to our next theorem.
This is sometimes written as
limx→a
f (g(x)) = limg(x)→M
f (g(x)).
Theorem 1.2.3 (Limit Composition Law) Suppose that limx→a
g(x) = M and
limx→M
f (x) = f (M). Then
limx→a
f (g(x)) = f (M).
calculus 15
Note the special form of the condition on f (x): it is not enough to know that
limx→L
f (x) exists, though it is a bit tricky to see why. Consider
f (x) =
3 if x = 2,
4 if x , 2.
and g(x) = 2. Now the conditions of Theorem 1.2.3 are not satisfied, and
limx→1
f (g(x)) = 3 but limx→2
f (x) = 4.
Many of the most familiar functions do satisfy the conditions of Theorem 1.2.3.
For example:
Theorem 1.2.4 (Limit Root Law) Suppose that n is a positive integer. Then
limx→a
n√
x =n√
a,
provided that a is positive if n is even.
This theorem is not too difficult to prove from the definition of limit.
16
Exercises for Section 1.2
(1) For each of the following limits, limx→a
f (x) = L, use a graphing device to find δ
such that 0 < |x − a| < δ implies that |f (x) − L | < ε where ε = .1.
(a) limx→2
(3x + 1) = 7
(b) limx→1
(x2+ 2) = 3
(c) limx→π
sin(x) = 0
(d) limx→0
tan(x) = 0
(e) limx→1
√3x + 1 = 2
(f) limx→−2
√1 − 4x = 3
➠
The next set of exercises are for advanced students and can be skipped on first
reading.
(2) Use the definition of limits to explain why limx→0
x sin
(1
x
)
= 0. Hint: Use the fact
that | sin(a)| < 1 for any real number a. ➠
(3) Use the definition of limits to explain why limx→4
(2x − 5) = 3. ➠
(4) Use the definition of limits to explain why limx→−3
(−4x − 11) = 1. ➠
(5) Use the definition of limits to explain why limx→−2
π = π. ➠
(6) Use the definition of limits to explain why limx→−2
x2 − 4
x + 2= −4. ➠
(7) Use the definition of limits to explain why limx→4
x3= 64. ➠
(8) Use the definition of limits to explain why limx→1
(x2+ 3x − 1) = 3. ➠
(9) Use the definition of limits to explain why limx→9
x − 9√
x − 3= 6. ➠
(10) Use the definition of limits to explain why limx→2
1
x=
1
2. ➠
calculus 17
1.3 Limit Laws
In this section, we present a handful of tools to compute many limits without
explicitly working with the definition of limit. Each of these could be proved directly
as we did in the previous section.
Theorem 1.3.1 (Limit Laws) Suppose that limx→a
f (x) = L, limx→a
g(x) = M, k is
some constant, and n is a positive integer.
Constant Law limx→a
kf (x) = k limx→a
f (x) = kL.
Sum Law limx→a
(f (x) + g(x)) = limx→a
f (x) + limx→a
g(x) = L +M.
Product Law limx→a
(f (x)g(x)) = limx→a
f (x) · limx→a
g(x) = LM.
Quotient Law limx→a
f (x)
g(x)=
limx→a f (x)
limx→a g(x)=
L
M, if M , 0.
Power Law limx→a
f (x)n=
(
limx→a
f (x)
)n
= Ln.
Root Law limx→a
n√
f (x) = n
√
limx→a
f (x) =n√
L provided if n is even, then f (x) ≥ 0
near a.
Composition Law If limx→a
g(x) = M and limx→M
f (x) = f (M), then limx→a
f (g(x)) =
f (M).
Roughly speaking, these rules say that to compute the limit of an algebraic
expression, it is enough to compute the limits of the “innermost bits” and then
combine these limits. This often means that it is possible to simply plug in a value
for the variable, since limx→a
x = a.
Example 1.3.2 Compute limx→1
x2 − 3x + 5
x − 2.
18
Solution Using limit laws,
limx→1
x2 − 3x + 5
x − 2=
limx→1 x2 − 3x + 5
limx→1(x − 2)
=limx→1 x2 − limx→1 3x + limx→1 5
limx→1 x − limx→1 2
=(limx→1 x)2 − 3 limx→1 x + 5
limx→1 x − 2
=12 − 3 · 1 + 5
1 − 2
=1 − 3 + 5
−1= −3.
It is worth commenting on the trivial limit limx→1
5. From one point of view this
might seem meaningless, as the number 5 can’t “approach” any value, since it is
simply a fixed number. But 5 can, and should, be interpreted here as the function
that has value 5 everywhere, f (x) = 5, with graph a horizontal line. From this
point of view it makes sense to ask what happens to the height of the function as x
approaches 1.
We’re primarily interested in limits that aren’t so easy, namely limits in which
a denominator approaches zero. The basic idea is to “divide out” by the offending
factor. This is often easier said than done—here we give two examples of algebraic
tricks that work on many of these limits.
Example 1.3.3 Compute limx→1
x2+ 2x − 3
x − 1.
Solution We can’t simply plug in x = 1 because that makes the denominator
zero. However, when taking limits we assume x , 1:
limx→1
x2+ 2x − 3
x − 1= lim
x→1
(x − 1)(x + 3)
x − 1
= limx→1
(x + 3) = 4
Limits allow us to examine functions where they arenot defined.
calculus 19
Example 1.3.4 Compute limx→−1
√x + 5 − 2
x + 1.
Solution Using limit laws,
limx→−1
√x + 5 − 2
x + 1= lim
x→−1
√x + 5 − 2
x + 1
√x + 5 + 2√
x + 5 + 2
= limx→−1
x + 5 − 4
(x + 1)(√
x + 5 + 2)
= limx→−1
x + 1
(x + 1)(√
x + 5 + 2)
= limx→−1
1√
x + 5 + 2=
1
4.
Here we are rationalizing the numerator by multiply-ing by the conjugate.
We’ll conclude with one more theorem that will allow us to compute more difficult
limits.
Theorem 1.3.5 (Squeeze Theorem) Suppose that g(x) ≤ f (x) ≤ h(x) for all
x close to a but not necessarily equal to a. If
limx→a
g(x) = L = limx→a
h(x),
then limx→a
f (x) = L.
For a nice discussion of this limit, see: Richman,Fred. A circular argument. College Math. J. 24(1993), no. 2, 160–162.
Example 1.3.6 Compute
limx→0
sin(x)
x.
20
Solution To compute this limit, use the Squeeze Theorem, Theorem 1.3.5. First
note that we only need to examine x ∈(−π
2,
π
2
)
and for the present time, we’ll
assume that x is positive—consider the diagrams below:
x
sin(x)
cos(x)u
v
x
1u
v
x
1
tan(x)
u
v
Triangle A Sector Triangle B
From our diagrams above we see that
Area of Triangle A ≤ Area of Sector ≤ Area of Triangle B
and computing these areas we find
cos(x) sin(x)
2≤
(x
2π
)
· π ≤tan(x)
2.
Multiplying through by 2, and recalling that tan(x) =sin(x)
cos(x)we obtain
cos(x) sin(x) ≤ x ≤sin(x)
cos(x).
Dividing through by sin(x) and taking the reciprocals, we find
cos(x) ≤sin(x)
x≤
1
cos(x).
calculus 21
Note, cos(−x) = cos(x) andsin(−x)
−x=
sin(x)
x, so these inequalities hold for all
x ∈(−π
2,
π
2
)
. Additionally, we know
limx→0
cos(x) = 1 = limx→0
1
cos(x),
and so we conclude by the Squeeze Theorem, Theorem 1.3.5, limx→0
sin(x)
x= 1.
22
Exercises for Section 1.3
Compute the limits. If a limit does not exist, explain why.
(1) limx→3
x2+ x − 12
x − 3 ➠
(2) limx→1
x2+ x − 12
x − 3 ➠
(3) limx→−4
x2+ x − 12
x − 3 ➠
(4) limx→2
x2+ x − 12
x − 2 ➠
(5) limx→1
√x + 8 − 3
x − 1 ➠
(6) limx→0+
√
1
x+ 2 −
√
1
x ➠
(7) limx→2
3 ➠
(8) limx→4
3x3 − 5x ➠
(9) limx→0
4x − 5x2
x − 1 ➠
(10) limx→1
x2 − 1
x − 1 ➠
(11) limx→0+
√2 − x2
x ➠
(12) limx→0+
√2 − x2
x + 1 ➠
(13) limx→a
x3 − a3
x − a ➠
(14) limx→2
(x2+ 4)3
➠
(15) limx→1
x − 5 if x , 1,
7 if x = 1.➠
calculus 23
1.4 Infinite Limits
Consider the function
f (x) =1
(x + 1)2
While the limx→−1
f (x) does not exist, see Figure 1.7, something can still be said.
−2 −1.5 −1 −0.5 0.5 1
20
40
60
80
100
x
y
Figure 1.7: A plot of f (x) =1
(x + 1)2.
Definition If f (x) grows arbitrarily large as x approaches a, we write
limx→a
f (x) = ∞
and say that the limit of f (x) approaches infinity as x goes to a.
If |f (x)| grows arbitrarily large as x approaches a and f (x) is negative, we
write
limx→a
f (x) = −∞
and say that the limit of f (x) approaches negative infinity as x goes to a.
On the other hand, if we consider the function
f (x) =1
(x − 1)
While we have limx→1
f (x) , ±∞, we do have one-sided limits, limx→1+
f (x) = ∞ and
limx→1−
f (x) = −∞, see Figure 1.8. −1 −0.5 0.5 1 1.5 2
−40
−20
20
40
x
y
Figure 1.8: A plot of f (x) =1
x − 1.
Definition If
limx→a
f (x) = ±∞, limx→a+
f (x) = ±∞, or limx→a−
f (x) = ±∞,
then the line x = a is a vertical asymptote of f (x).
24
Example 1.4.1 Find the vertical asymptotes of
f (x) =x2 − 9x + 14
x2 − 5x + 6.
Solution Start by factoring both the numerator and the denominator:
x2 − 9x + 14
x2 − 5x + 6=
(x − 2)(x − 7)
(x − 2)(x − 3)
Using limits, we must investigate when x → 2 and x → 3. Write
limx→2
(x − 2)(x − 7)
(x − 2)(x − 3)= lim
x→2
(x − 7)
(x − 3)
=−5
−1
= 5.
Now write
limx→3
(x − 2)(x − 7)
(x − 2)(x − 3)= lim
x→3
(x − 7)
(x − 3)
= limx→3
−4
x − 3.
Since limx→3+
x − 3 approaches 0 from the right and the numerator is negative,
limx→3+
f (x) = −∞. Since limx→3−
x − 3 approaches 0 from the left and the numerator
is negative, limx→3−
f (x) = ∞. Hence we have a vertical asymptote at x = 3, see
Figure 1.9.
1.5 2 2.5 3 3.5 4
−40
−20
20
40
x
y
Figure 1.9: A plot of f (x) =x2 − 9x + 14
x2 − 5 + 6.
calculus 25
Exercises for Section 1.4
Compute the limits. If a limit does not exist, explain why.
(1) limx→1−
1
x2 − 1 ➠
(2) limx→4−
3
x2 − 2 ➠
(3) limx→−1+
1 + 2x
x3 − 1 ➠
(4) limx→3+
x − 9
x2 − 6x + 9 ➠
(5) limx→5
1
(x − 5)4 ➠
(6) limx→−2
1
(x2 + 3x + 2)2 ➠
(7) limx→0
1xx5 − cos(x) ➠
(8) limx→0+
x − 11
sin(x) ➠
(9) Find the vertical asymptotes of
f (x) =x − 3
x2 + 2x − 3.
➠
(10) Find the vertical asymptotes of
f (x) =x2 − x − 6
x + 4.
➠
26
1.5 Limits at Infinity
Consider the function:
f (x) =6x − 9
x − 1
As x approaches infinity, it seems like f (x) approaches a specific value. This is a
0.5 1 1.5 2 2.5 3
−10
10
20
x
y
Figure 1.10: A plot of f (x) =6x − 9
x − 1.
limit at infinity.
Definition If f (x) becomes arbitrarily close to a specific value L by making x
sufficiently large, we write
limx→∞
f (x) = L
and we say, the limit at infinity of f (x) is L.
If f (x) becomes arbitrarily close to a specific value L by making x sufficiently
large and negative, we write
limx→−∞
f (x) = L
and we say, the limit at negative infinity of f (x) is L.
Example 1.5.1 Compute
limx→∞
6x − 9
x − 1.
Solution Write
limx→∞
6x − 9
x − 1= lim
x→∞
6x − 9
x − 1
1/x
1/x
= limx→∞
6xx− 9
x
xx− 1
x
= limx→∞
6
1
= 6.
Here is a somewhat different example of a limit at infinity.
calculus 27
Example 1.5.2 Compute
limx→∞
sin(7x)
x+ 4.
5 10 15 20
3.5
4
4.5
5
x
y
Figure 1.11: A plot of f (x) =sin(7x)
x+ 4.
Solution We can bound our function
−1/x + 4 ≤ sin(7x)
x+ 4 ≤ 1/x + 4.
Since
limx→∞−1/x + 4 = 4 = lim
x→∞1/x + 4
we conclude by the Squeeze Theorem, Theorem 1.3.5, limx→∞
sin(7x)
x+ 4 = 4.
Definition If
limx→∞
f (x) = L or limx→−∞
f (x) = L,
then the line y = L is a horizontal asymptote of f (x).
Example 1.5.3 Give the horizontal asymptotes of
f (x) =6x − 9
x − 1
Solution From our previous work, we see that limx→∞
f (x) = 6, and upon further
inspection, we see that limx→−∞
f (x) = 6. Hence the horizontal asymptote of f (x) is
the line y = 6.
It is a common misconception that a function cannot cross an asymptote. As
the next example shows, a function can cross an asymptote, and in this case this
occurs an infinite number of times!
Example 1.5.4 Give a horizontal asymptote of
f (x) =sin(7x)
x+ 4.
28
Solution Again from previous work, we see that limx→∞
f (x) = 4. Hence y = 4 is
a horizontal asymptote of f (x).
We conclude with an infinite limit at infinity.
Example 1.5.5 Compute
limx→∞
ln(x)
5 10 15 20
−1
1
2
3
4
x
y
Figure 1.12: A plot of f (x) = ln(x).
Solution The function ln(x) grows very slowly, and seems like it may have a
horizontal asymptote, see Figure 1.12. However, if we consider the definition of
the natural log
ln(x) = y ⇔ ey= x
Since we need to raise e to higher and higher values to obtain larger numbers,
we see that ln(x) is unbounded, and hence limx→∞
ln(x) = ∞.
calculus 29
Exercises for Section 1.5
Compute the limits.
(1) limx→∞
1
x ➠
(2) limx→∞
−x√
4 + x2 ➠
(3) limx→∞
2x2 − x + 1
4x2 − 3x − 1 ➠
(4) limx→−∞
x3 − 4
3x2 + 4x − 1 ➠
(5) limx→∞
(4
x+ π
)
➠
(6) limx→∞
cos(x)
ln(x) ➠
(7) limx→∞
sin(
x7)
√x
➠
(8) limx→∞
(
17 +32
x−
(sin(x/2))2
x3
)
➠
(9) Suppose a population of feral cats on a certain college campus t years from now
is approximated by
p(t) =1000
5 + 2e−0.1t.
Approximately how many feral cats are on campus 10 years from now? 50 years
from now? 100 years from now? 1000 years from now? What do you notice
about the prediction—is this realistic? ➠
(10) The amplitude of an oscillating spring is given by
a(t) =sin(t)
t.
What happens to the amplitude of the oscillation over a long period of time?
➠
30
1.6 Continuity
Informally, a function is continuous if you can “draw it” without “lifting your pencil.”
We need a formal definition.
Definition A function f is continuous at a point a if limx→a
f (x) = f (a).
2 4 6 8 10
1
2
3
4
5
x
y
Figure 1.13: A plot of a function with discontinuitiesat x = 4 and x = 6.
Example 1.6.1 Find the discontinuities (the values for x where a function is
not continuous) for the function given in Figure 1.13.
Solution From Figure 1.13 we see that limx→4
f (x) does not exist as
limx→4−
f (x) = 1 and limx→4+
f (x) ≈ 3.5
Hence limx→4
f (x) , f (4), and so f (x) is not continuous at x = 4.
We also see that limx→6
f (x) ≈ 3 while f (6) = 2. Hence limx→6
f (x) , f (6), and so
f (x) is not continuous at x = 6.
Building from the definition of continuous at a point, we can now define what it
means for a function to be continuous on an interval.
Definition A function f is continuous on an interval if it is continuous at
every point in the interval.
In particular, we should note that if a function is not defined on an interval, then
it cannot be continuous on that interval.
−0.2 −0.1 0.1 0.2x
y
Figure 1.14: A plot of
f (x) =
5√x sin
(1
x
)
if x , 0,
0 if x = 0.
Example 1.6.2 Consider the function
f (x) =
5√
x sin
(1
x
)
if x , 0,
0 if x = 0,
see Figure 1.14. Is this function continuous?
calculus 31
Solution Considering f (x), the only issue is when x = 0. We must show that
limx→0
f (x) = 0. Note
−| 5√
x | ≤ f (x) ≤ | 5√
x |.
Since
limx→0−| 5√
x | = 0 = limx→0| 5√
x |,
we see by the Squeeze Theorem, Theorem 1.3.5, that limx→0
f (x) = 0. Hence f (x) is
continuous.
Here we see how the informal definition of continuity being that you can
“draw it” without “lifting your pencil” differs from the formal definition.
We close with a useful theorem about continuous functions:
Theorem 1.6.3 (Intermediate Value Theorem) If f (x) is a function that is
continuous for all x in the closed interval [a, b] and d is between f (a) and
f (b), then there is a number c in [a, b] such that f (c) = d.
The Intermediate Value Theorem is most frequentlyused when d = 0.
For a nice proof of this theorem, see: Walk, StephenM. The intermediate value theorem is NOT obvious—
and I am going to prove it to you. College Math. J. 42(2011), no. 4, 254–259.In Figure 1.15, we see a geometric interpretation of this theorem.
Example 1.6.4 Explain why the function f (x) = x3+ 3x2
+ x − 2 has a root
between 0 and 1.
Solution By Theorem 1.3.1, limx→a
f (x) = f (a), for all real values of a, and hence
f is continuous. Since f (0) = −2 and f (1) = 3, and 0 is between −2 and 3, there
is a c ∈ [0, 1] such that f (c) = 0.
This example also points the way to a simple method for approximating roots.
Example 1.6.5 Approximate a root of f (x) = x3+ 3x2
+ x − 2 to one decimal
place.
Solution If we compute f (0.1), f (0.2), and so on, we find that f (0.6) < 0 and
f (0.7) > 0, so by the Intermediate Value Theorem, f has a root between 0.6
32
a c b
f (a)
f (c) = d
f (b)
x
y Figure 1.15: A geometric interpretation of the In-termediate Value Theorem. The function f (x) iscontinuous on the interval [a, b]. Since d is in theinterval [f (a), f (b)], there exists a value c in [a, b]
such that f (c) = d.
and 0.7. Repeating the process with f (0.61), f (0.62), and so on, we find that
f (0.61) < 0 and f (0.62) > 0, so f has a root between 0.61 and 0.62, and the
root is 0.6 rounded to one decimal place.
calculus 33
Exercises for Section 1.6
(1) Consider the function
f (x) =√
x − 4
Is f (x) continuous at the point x = 4? Is f (x) a continuous function on R? ➠
(2) Consider the function
f (x) =1
x + 3
Is f (x) continuous at the point x = 3? Is f (x) a continuous function on R? ➠
(3) Consider the function
f (x) =
2x − 3 if x < 1,
0 if x ≥ 1.
Is f (x) continuous at the point x = 1? Is f (x) a continuous function on R? ➠
(4) Consider the function
f (x) =
x2+ 10x + 25
x − 5if x , 5,
10 if x = 5.
Is f (x) continuous at the point x = 5? Is f (x) a continuous function on R? ➠
(5) Consider the function
f (x) =
x2+ 10x + 25
x + 5if x , −5,
0 if x = −5.
Is f (x) continuous at the point x = −5? Is f (x) a continuous function on R?
➠
(6) Determine the interval(s) on which the function f (x) = x7+ 3x5 − 2x + 4 is
continuous. ➠
34
(7) Determine the interval(s) on which the function f (x) =x2 − 2x + 1
x + 4is continuous.
➠
(8) Determine the interval(s) on which the function f (x) =1
x2 − 9is continuous.
➠
(9) Approximate a root of f (x) = x3 − 4x2+ 2x + 2 to two decimal places. ➠
(10) Approximate a root of f (x) = x4+ x3 − 5x + 1 to two decimal places. ➠
2 Basics of Derivatives
2.1 Slopes of Tangent Lines via Limits
Suppose that f (x) is a function. It is often useful to know how sensitive the value of
f (x) is to small changes in x. To give you a feeling why this is true, consider the
following:
• If p(t) represents the position of an object with respect to time, the rate of change
gives the velocity of the object.
• If v(t) represents the velocity of an object with respect to time, the rate of change
gives the acceleration of the object.
• The rate of change of a function can help us approximate a complicated function
with a simple function.
• The rate of change of a function can be used to help us solve equations that we
would not be able to solve via other methods.
The rate of change of a function is the slope of the tangent line. Part of our goal
will be to give a formal definition of a tangent line. For now, consider the following
informal definition:
Given a function f (x), if one can “zoom in” on f (x) sufficiently so that f (x) seems to be
a straight line, then that line is the tangent line to f (x) at the point determined by x.
While this is merely an informal definition of a tangent line, it contains the essence
of how the formal definition will be constructed. We illustrate this informal definition
with Figure 2.1.
36
x
y
Figure 2.1: Given a function f (x), if one can “zoomin” on f (x) sufficiently so that f (x) seems to be astraight line, then that line is the tangent line tof (x) at the point determined by x.
The derivative of a function f (x) at x, is the slope of the tangent line at x. To find
the slope of this line, we consider secant lines, lines that locally intersect the curve
at two points. The slope of any secant line that passes through the points (x, f (x))
and (x + h, f (x + h)) is given by
∆y
∆x=
f (x + h) − f (x)
(x + h) − x=
f (x + h) − f (x)
h,
see Figure 2.2. This leads to the limit definition of the derivative:
Definition of the Derivative The derivative of f (x) is the function
d
dxf (x) = lim
h→0
f (x + h) − f (x)
h.
If this limit does not exist for a given value of x, then f (x) is not differentiable
at x.
x x + h
f (x)
f (x + h)
x
y
Figure 2.2: Tangent lines can be found as the limitof secant lines. The slope of the tangent line is given
by limh→0
f (x + h) − f (x)
h.
calculus 37
Definition There are several different notations for the derivative, we’ll mainly
used
dxf (x) = f ′(x).
If one is working with a function of a variable other than x, say t we write
d
dtf (t) = f ′(t).
However, if y = f (x),dy
dx, y, and Dx f (x) are also used.
Now we will give a number of examples, starting with a basic example.
Example 2.1.1 Computed
dx(x3+ 1).
Solution Using the definition of the derivative,
d
dxf (x) = lim
h→0
(x + h)3+ 1 − (x3
+ 1)
h
= limh→0
x3+ 3x2h + 3xh2
+ h3+ 1 − x3 − 1
h
= limh→0
3x2h + 3xh2+ h3
h
= limh→0
(3x2+ 3xh + h2)
= 3x2.
See Figure 2.3.
−1.5 −1 −0.5 0.5 1
−4
−2
2
4
f (x)
f ′(x)
x
y
Figure 2.3: A plot of f (x) = x3+ 1 and f ′(x) = 3x2.
Next we will consider the derivative a function that is not continuous on R.
Example 2.1.2 Computed
dt
1
t.
38
Solution Using the definition of the derivative,
d
dt
1
t= lim
h→0
1t+h− 1
t
h
= limh→0
tt(t+h)
− t+ht(t+h)
h
= limh→0
t−(t+h)
t(t+h)
h
= limh→0
t − t − h
t(t + h)h
= limh→0
−h
t(t + h)h
= limh→0
−1
t(t + h)
=−1
t2.
This function is differentiable at all real numbers except for t = 0, see Figure 2.4.
−3 −2 −1 1 2 3
−4
−2
2
4
f (t)
f ′(t)
t
y
Figure 2.4: A plot of f (t) =1
tand f ′(t) =
−1
t2.
As you may have guessed, there is some connection to continuity and differentia-
bility.
Theorem 2.1.3 (Differentiability implies Continuity) If f (x) is a differen-
tiable function at x = a, then f (x) is continuous at x = a.
Proof We want to show that f (x) is continuous at x = a, hence we must show
that
limx→a
f (x) = f (a).
calculus 39
Consider
limx→a
(f (x) − f (a)) = limx→a
(
(x − a)f (x) − f (a)
x − a
)
Multiply and divide by (x − a).
= limh→0
h · f (a + h) − f (a)
hSet x = a + h.
=
(
limh→0
h
) (
limh→0
f (a + h) − f (a)
h
)
Limit Law.
= 0 · f ′(a) = 0.
Since
limx→a
(f (x) − f (a)) = 0
we see that limx→a
f (x) = f (a), and so f (x) is continuous.
This theorem is often written as its contrapositive:
If f (x) is not continuous at x = a, then f (x) is not differentiable at x = a.
Let’s see a function that is continuous whose derivative does not exist everywhere.
Example 2.1.4 Computed
dx|x |.
−3 −2 −1 1 2 3
−2
−1
1
2
3
f (t)
f ′(t)
x
y
Figure 2.5: A plot of f (x) = |x | and
f ′(x) =
1 if x > 0,
−1 if x < 0.
Solution Using the definition of the derivative,
d
dx|x | = lim
h→0
|x + h | − |x |h
.
If x is positive we may assume that x is larger than h, as we are taking the limit
as h goes to 0,
limh→0
|x + h | − |x |h
= limh→0
x + h − x
h
= limh→0
h
h
= 1.
40
If x is negative we may assume that |x | is larger than h, as we are taking the
limit as h goes to 0,
limh→0
|x + h | − |x |h
= limh→0
x − h − x
h
= limh→0+
−h
h
= −1.
However we still have one case left, when x = 0. In this situation, we must
consider the one-sided limits:
limh→0+
|x + h | − |x |h
and limh→0−
|x + h | − |x |h
.
In the first case,
limh→0+
|x + h | − |x |h
= limh→0+
0 + h − 0
h
= limh→0+
h
h
= 1.
On the other hand
limh→0−
|x + h | − |x |h
= limh→0−
|0 + h | − 0
h
= limh→0−
|h |h
= −1.
Hence we see that the derivative is
f ′(x) =
1 if x > 0,
−1 if x < 0.
Note this function is undefined at 0, see Figure 2.5.
Thus from Theorem 2.1.3, we see that all differentiable functions on R are
continuous on R. Nevertheless as the previous example shows, there are continuous
calculus 41
functions on R that are not differentiable on R.
42
Exercises for Section 2.1
These exercises are conceptual in nature and require one to think about what the
derivative means.
(1) If the line y = 7x − 4 is tangent to f (x) at x = 2, find f (2) and f ′(2). ➠
(2) Here are plots of four functions.
1 2
−4
−2
2
4
x
y
1 2
−4
−2
2
4
x
y
−1 1 2 3
−4
−2
2
4
x
y
1 2
−4
−2
2
4
x
y
p(x) q(x) r(x) s(x)
Two of these functions are the derivatives of the other two, identify which
functions are the derivatives of the others. ➠
(3) If f (3) = 6 and f (3.1) = 6.4, estimate f ′(3). ➠
(4) If f (−2) = 4 and f (−2 + h) = (h + 2)2, compute f ′(−2). ➠
(5) If f ′(x) = x3 and f (1) = 2, approximate f (1.2). ➠
1 2 3 4 5 6
−1
1
2
3
4
x
y
Figure 2.6: A plot of f (x).
(6) Consider the plot of f (x) in Figure 2.6.
(a) On which subinterval(s) of [0, 6] is f (x) continuous?
(b) On which subinterval(s) of [0, 6] is f (x) differentiable?
(c) Sketch a plot of f ′(x).
➠
calculus 43
These exercises are computational in nature.
(7) Let f (x) = x2 − 4. Use the definition of the derivative to compute f ′(−3) and find
the equation of the tangent line to the curve at x = −3. ➠
(8) Let f (x) =1
x + 2. Use the definition of the derivative to compute f ′(1) and find
the equation of the tangent line to the curve at x = 1. ➠
(9) Let f (x) =√
x − 3. Use the definition of the derivative to compute f ′(5) and find
the equation of the tangent line to the curve at x = 5. ➠
(10) Let f (x) =1√
x. Use the definition of the derivative to compute f ′(4) and find
the equation of the tangent line to the curve at x = 4. ➠
44
2.2 Basic Derivative Rules
It is tedious to compute a limit every time we need to know the derivative of a
function. Fortunately, we can develop a small collection of examples and rules that
allow us to compute the derivative of almost any function we are likely to encounter.
We will start simply and build-up to more complicated examples.
The Constant Rule
The simplest function is a constant function. Recall that derivatives measure the
rate of change of a function at a given point. Hence, the derivative of a constant
function is zero. For example:
• The constant function plots a horizontal line—so the slope of the tangent line is 0.
• If p(t) represents the position of an object with respect to time and p(t) is constant,
then the object is not moving, so its velocity is zero. Henced
dtp(t) = 0.
• If v(t) represents the velocity of an object with respect to time and v(t) is constant,
then the object’s acceleration is zero. Henced
dtv(t) = 0.
The examples above lead us to our next theorem. To gain intuition, you should compute the derivativeof f (x) = 6 using the limit definition of the derivative.
Theorem 2.2.1 (The Constant Rule) Given a constant c,
d
dxc = 0.
Proof From the limit definition of the derivative, write
d
dxc = lim
h→0
c − c
h
= limh→0
0
h
= limh→0
0 = 0.
calculus 45
The Power Rule
Now let’s examine derivatives of powers of a single variable. Here we have a nice
rule. To gain intuition, you should compute the deriva-tive of f (x) = x3 using the limit definition of thederivative.
Theorem 2.2.2 (The Power Rule) For any real number n,
d
dxxn= nxn−1.
Recall, the Binomial Theorem states that if n is anonnegative integer, then
(a+b)n= anb0
+
(
n
1
)
an−1b1+· · ·+
(
n
n − 1
)
a1bn−1+a0bn
where (
n
k
)
=n!
k!(n − k)!.
Proof At this point we will only prove this theorem for n being a positive integer.
Later in Section 5.3, we will give the complete proof. From the limit definition of
the derivative, writed
dxxn= lim
h→0
(x + h)n − xn
h.
Start by expanding the term (x + h)n
d
dxxn= lim
h→0
xn+
(n1
)
xn−1h +(n2
)
xn−2h2+ · · · +
(n
n−1
)
xhn−1+ hn − xn
h
Note, by the Binomial Theorem, we write
(
n
k
)
for the coefficients. Canceling the
terms xn and −xn , and noting
(
n
1
)
=
(
n
n − 1
)
= n, write
d
dxxn= lim
h→0
nxn−1h +(n2
)
xn−2h2+ · · · +
(n
n−1
)
xhn−1+ hn
h
= limh→0
nxn−1+
(
n
2
)
xn−2h + · · · +(
n
n − 1
)
xhn−2+ hn−1.
Since every term but the first has a factor of h, we see
d
dxxn= lim
h→0
(x + h)n − xn
h= nxn−1.
Now we will show you several examples. We begin with something basic.
46
Example 2.2.3 Computed
dxx13.
Solution Applying the power rule, we write
d
dxx13= 13x12.
Sometimes, it is not as obvious that one should apply the power rule.
Example 2.2.4 Computed
dx
1
x4.
Solution Applying the power rule, we write
d
dx
1
x4=
d
dxx−4= −4x−5.
The power rule also applies to radicals once we rewrite them as exponents.
Example 2.2.5 Computed
dx
5√
x.
Solution Applying the power rule, we write
d
dx
5√
x =d
dxx1/5
=x−4/5
5.
The Sum Rule
We want to be able to take derivatives of functions “one piece at a time.” The sum
rule allows us to do this. The sum rule says that we can add the rates of change
of two functions to obtain the rate of change of the sum of both functions. For
example, viewing the derivative as the velocity of an object, the sum rule states that
the velocity of the person walking on a moving bus is the sum of the velocity of the
bus and the walking person.
calculus 47
Theorem 2.2.6 (The Sum Rule) If f (x) and g(x) are differentiable and c is
a constant, then
(a)d
dx
(
f (x) + g(x))
= f ′(x) + g′(x),
(b)d
dx
(
f (x) − g(x))
= f ′(x) − g′(x),
(c)d
dx
(
c · f (x))
= c · f ′(x).
f (x)
g(x)
f (x) + g(x)
f ′(a)h
g′(a)h
f ′(a)h + g′(a)h
︸ ︷︷ ︸
h
ax
y
Figure 2.7: A geometric interpretation of the sumrule. Since every point on f (x)+g(x) is the sum of thecorresponding points on f (x) and g(x), increasing a
by a “small amount” h, increases f (a) + g(a) by thesum of f ′(a)h and g′(a)h. Hence,
∆y
∆x≈
f ′(a)h + g′(a)h
h= f ′(a) + g′(a).
Proof We will only prove part (a) above, the rest are similar. Write
d
dx
(
f (x) + g(x))
= limh→0
f (x + h) + g(x + h) − (f (x) + g(x))
h
= limh→0
f (x + h) + g(x + h) − f (x) − g(x)
h
= limh→0
f (x + h) − f (x) + g(x + h) − g(x)
h
= limh→0
(
f (x + h) − f (x)
h+
g(x + h) − g(x)
h
)
= limh→0
f (x + h) − f (x)
h+ lim
h→0
g(x + h) − g(x)
h
= f ′(x) + g′(x).
Example 2.2.7 Computed
dx
(
x5+
1
x
)
.
Solution Write
d
dx
(
x5+
1
x
)
=d
dxx5+
d
dxx−1
= 5x4 − x−2.
48
Example 2.2.8 Compute
d
dx
(
33√
x− 2√
x +1
x7
)
.
Solution Write
d
dx
(
33√
x− 2√
x +1
x7
)
= 3d
dxx−1/3 − 2
d
dxx1/2
+d
dxx−7
= −x−4/3 − x−1/2 − 7x−8.
The Derivative of ex
We don’t know anything about derivatives that allows us to compute the derivatives
of exponential functions without getting our hands dirty. Let’s do a little work with
the definition of the derivative:
d
dxax= lim
h→0
ax+h − ax
h
= limh→0
axah − ax
h
= limh→0
ax ah − 1
h
= ax limh→0
ah − 1
h
= ax · (constant)︸ ︷︷ ︸
limh→0ah−1
h
There are two interesting things to note here: We are left with a limit that involves h
but not x, which means that whatever limh→0
(ah −1)/h is, we know that it is a number,
that is, a constant. This means that ax has a remarkable property: Its derivative is
a constant times itself. Unfortunately it is beyond the scope of this text to compute
the limit
limh→0
ah − 1
h.
However, we can look at some examples. Consider (2h − 1)/h and (3h − 1)/h:
calculus 49
h (2h − 1)/h
−1 .5
−0.1 ≈ 0.6700
−0.01 ≈ 0.6910
−0.001 ≈ 0.6929
−0.0001 ≈ 0.6931
−0.00001 ≈ 0.6932
h (2h − 1)/h
1 1
0.1 ≈ 0.7177
0.01 ≈ 0.6956
0.001 ≈ 0.6834
0.0001 ≈ 0.6932
0.00001 ≈ 0.6932
h (3h − 1)/h
−1 ≈ 0.6667
−0.1 ≈ 1.0404
−0.01 ≈ 1.0926
−0.001 ≈ 1.0980
−0.0001 ≈ 1.0986
−0.00001 ≈ 1.0986
h (3h − 1)/h
1 2
0.1 ≈ 1.1612
0.01 ≈ 1.1047
0.001 ≈ 1.0992
0.0001 ≈ 1.0987
0.00001 ≈ 1.0986
While these tables don’t prove a pattern, it turns out that
limh→0
2h − 1
h≈ .7 and lim
h→0
3h − 1
h≈ 1.1.
Moreover, if you do more examples you will find that the limit varies directly with
the value of a: bigger a, bigger limit; smaller a, smaller limit. As we can already
see, some of these limits will be less than 1 and some larger than 1. Somewhere
between a = 2 and a = 3 the limit will be exactly 1. This happens when
a = e = 2.718281828459045 . . . .
This brings us to our next definition.
Definition Euler’s number is defined to be the number e such that
limh→0
eh − 1
h= 1.
Now we see that the function ex has a truly remarkable property:
Theorem 2.2.9 (The Derivative of ex)
d
dxex= ex .
50
Proof From the limit definition of the derivative, write
d
dxex= lim
h→0
ex+h − ex
h
= limh→0
exeh − ex
h
= limh→0
ex eh − 1
h
= ex limh→0
eh − 1
h
= ex .
Hence ex is its own derivative. In other words, the slope of the plot of ex is the
same as its height, or the same as its second coordinate: The function f (x) = ex
goes through the point (a, ea) and has slope ea there, no matter what a is.
Example 2.2.10 Compute:
d
dx
(
8√
x + 7ex)
Solution Write:
d
dx
(
8√
x + 7ex)
= 8d
dxx1/2
+ 7d
dxex
= 4x−1/2+ 7ex .
calculus 51
Exercises for Section 2.2
Compute:
(1)d
dx5 ➠
(2)d
dx− 7 ➠
(3)d
dxe7
➠
(4)d
dx
1√
2➠
(5)d
dxx100
➠
(6)d
dxx−100
➠
(7)d
dx
1
x5 ➠
(8)d
dxxπ
➠
(9)d
dxx3/4
➠
(10)d
dx
1
( 7√
x)9 ➠
(11)d
dx
(
5x3+ 12x2 − 15
)
➠
(12)d
dx
(
−4x5+ 3x2 −
5
x2
)
➠
(13)d
dx5(−3x2
+ 5x + 1) ➠
(14)d
dx
(
3√
x +1
x− xe
)
➠
(15)d
dx
(
x2
x7+
√x
x
)
➠
Expand or simplify to compute the following:
(16)d
dx
(
(x + 1)(x2+ 2x − 3)
)
➠
(17)d
dx
x3 − 2x2 − 5x + 6
(x − 1) ➠
(18)d
dx
x − 5√
x −√
5➠
(19)d
dx((x + 1)(x + 1)(x − 1)(x − 1)) ➠
(20) Suppose the position of an object at time t is given by f (t) = −49t2/10 + 5t + 10.
Find a function giving the velocity of the object at time t. The acceleration of an
object is the rate at which its velocity is changing, which means it is given by
the derivative of the velocity function. Find the acceleration of the object at time
t. ➠
52
(21) Let f (x) = x3 and c = 3. Sketch the graphs of f (x), cf (x), f ′(x), and (cf (x))′ on
the same diagram. ➠
(22) Find a cubic polynomial whose graph has horizontal tangents at (−2, 5) and
(2, 3). ➠
(23) Find an equation for the tangent line to f (x) = x3/4 − 1/x at x = −2. ➠
(24) Find an equation for the tangent line to f (x) = 3x2 − π3 at x = 4. ➠
(25) Prove thatd
dx(cf (x)) = cf ′(x) using the definition of the derivative. ➠
3 Curve Sketching
Whether we are interested in a function as a purely mathematical object or in
connection with some application to the real world, it is often useful to know what
the graph of the function looks like. We can obtain a good picture of the graph
using certain crucial information provided by derivatives of the function and certain
limits.
3.1 Extrema
Local extrema on a function are points on the graph where the y coordinate is larger
(or smaller) than all other y coordinates on the graph at points “close to” (x, y).
Definition
(a) A point (x, f (x)) is a local maximum if there is an interval a < x < b with
f (x) ≥ f (z) for every z in (a, b).
(b) A point (x, f (x)) is a local minimum if there is an interval a < x < b with
f (x) ≤ f (z) for every z in (a, b).
A local extremum is either a local maximum or a local minimum.
Local maximum and minimum points are quite distinctive on the graph of a
function, and are therefore useful in understanding the shape of the graph. In
many applied problems we want to find the largest or smallest value that a function
54
achieves (for example, we might want to find the minimum cost at which some task
can be performed) and so identifying maximum and minimum points will be useful
for applied problems as well.
If (x, f (x)) is a point where f (x) reaches a local maximum or minimum, and if the
derivative of f exists at x, then the graph has a tangent line and the tangent line
must be horizontal. This is important enough to state as a theorem, though we will
not prove it.
Theorem 3.1.1 (Fermat’s Theorem) If f (x) has a local extremum at x = a
and f (x) is differentiable at a, then f ′(a) = 0.
−0.5 0.5 1 1.5 2 2.5 3
−2
2
f (x)
f ′(x)
x
y
Figure 3.1: A plot of f (x) = x3 − 4x2+ 3x and f ′(x) =
3x2 − 8x + 3.
Thus, the only points at which a function can have a local maximum or minimum are
points at which the derivative is zero, see Figure 3.1, or the derivative is undefined,
as in Figure 3.2. This brings us to our next definition.
−3 −2 −1 1 2 3
−2
−1
1
2
f (x)
f ′(x)
x
y
Figure 3.2: A plot of f (x) = x2/3 and f ′(x) =2
3x1/3.
Definition Any value of x for which f ′(x) is zero or undefined is called a
critical point for f (x).
Warning When looking for local maximum and minimum points, you are likely
to make two sorts of mistakes:
• You may forget that a maximum or minimum can occur where the deriva-
tive does not exist, and so forget to check whether the derivative exists
everywhere.
• You might assume that any place that the derivative is zero is a local
maximum or minimum point, but this is not true, see Figure 3.3.
Since the derivative is zero or undefined at both local maximum and local
minimum points, we need a way to determine which, if either, actually occurs. The
most elementary approach is to test directly whether the y coordinates near the
calculus 55
potential maximum or minimum are above or below the y coordinate at the point of
interest.
−1 −0.5 0.5 1
−2
2
f (x)f ′(x)
x
y
Figure 3.3: A plot of f (x) = x3 and f ′(x) = 3x2. Whilef ′(0) = 0, there is neither a maximum nor minimumat (0, f (0)).
It is not always easy to compute the value of a function at a particular point. The
task is made easier by the availability of calculators and computers, but they have
their own drawbacks—they do not always allow us to distinguish between values
that are very close together. Nevertheless, because this method is conceptually
simple and sometimes easy to perform, you should always consider it.
Example 3.1.2 Find all local maximum and minimum points for the function
f (x) = x3 − x.
Solution Writed
dxf (x) = 3x2 − 1.
This is defined everywhere and is zero at x = ±√
3/3. Looking first at x =√
3/3,
we see that
f (√
3/3) = −2√
3/9.
Now we test two points on either side of x =√
3/3, making sure that neither is
farther away than the nearest critical point; since√
3 < 3,√
3/3 < 1 and we
can use x = 0 and x = 1. Since
f (0) = 0 > −2√
3/9 and f (1) = 0 > −2√
3/9,
there must be a local minimum at x =√
3/3.
For x = −√
3/3, we see that f (−√
3/3) = 2√
3/9. This time we can use x = 0
and x = −1, and we find that f (−1) = f (0) = 0 < 2√
3/9, so there must be a
local maximum at x = −√
3/3, see Figure 3.4.
−1.5 −1 −0.5 0.5 1 1.5
−2
−1
1
2
f (x)f ′(x)
x
y
Figure 3.4: A plot of f (x) = x3−x and f ′(x) = 3x2−1.
56
Exercises for Section 3.1
In the following problems, find the x values for local maximum and minimum points
by the method of this section.
(1) y = x2 − x ➠
(2) y = 2 + 3x − x3➠
(3) y = x3 − 9x2+ 24x ➠
(4) y = x4 − 2x2+ 3 ➠
(5) y = 3x4 − 4x3➠
(6) y = (x2 − 1)/x ➠
(7) y = −x4
4+ x3
+ x2➠
(8) f (x) =
x − 1 x < 2
x2 x ≥ 2➠
(9) f (x) =
x − 3 x < 3
x3 3 ≤ x ≤ 5
1/x x > 5
➠
(10) f (x) = x2 − 98x + 4 ➠
(11) f (x) =
−2 x = 0
1/x2 x , 0➠
(12) How many critical points can a quadratic polynomial function have? ➠
(13) Explore the family of functions f (x) = x3+ cx + 1 where c is a constant. How
many and what types of local extrema are there? Your answer should depend
on the value of c, that is, different values of c will give different answers. ➠
calculus 57
3.2 The First Derivative Test
The method of the previous section for deciding whether there is a local maximum
or minimum at a critical point by testing “near-by” points is not always convenient.
Instead, since we have already had to compute the derivative to find the critical
points, we can use information about the derivative to decide. Recall that
• If f ′(x) > 0 on an interval, then f (x) is increasing on that interval.
• If f ′(x) < 0 on an interval, then f (x) is decreasing on that interval.
So how exactly does the derivative tell us whether there is a maximum, minimum,
or neither at a point? Use the first derivative test.
Theorem 3.2.1 (First Derivative Test) Suppose that f (x) is continuous on
an interval, and that f ′(a) = 0 for some value of a in that interval.
• If f ′(x) > 0 to the left of a and f ′(x) < 0 to the right of a, then f (a) is a
local maximum.
• If f ′(x) < 0 to the left of a and f ′(x) > 0 to the right of a, then f (a) is a
local minimum.
• If f ′(x) has the same sign to the left and right of f ′(a), then f ′(a) is not a
local extremum.
Example 3.2.2 Consider the function
f (x) =x4
4+
x3
3− x2
Find the intervals on which f (x) is increasing and decreasing and identify the
local extrema of f (x).
Solution Start by computing
d
dxf (x) = x3
+ x2 − 2x.
58
Now we need to find when this function is positive and when it is negative. To
do this, solve
f ′(x) = x3+ x2 − 2x = 0.
Factor f ′(x)
f ′(x) = x3+ x2 − 2x
= x(x2+ x − 2)
= x(x + 2)(x − 1).
So the critical points (when f ′(x) = 0) are when x = −2, x = 0, and x = 1. Now
we can check points between the critical points to find when f ′(x) is increasing
and decreasing:
f ′(−3) = −12 f ′(.5) = −0.625 f ′(−1) = 2 f ′(2) = 8
From this we can make a sign table:
−2 0 1
f ′(x) < 0 f ′(x) < 0f ′(x) > 0 f ′(x) > 0
Decreasing DecreasingIncreasing Increasing
Hence f (x) is increasing on (−2, 0) ∪ (1,∞) and f (x) is decreasing on
(−∞,−2) ∪ (0, 1). Moreover, from the first derivative test, Theorem 3.2.1, the
local maximum is at x = 0 while the local minima are at x = −2 and x = 1, see
Figure 3.5.
−3 −2 −1 1 2
−4
−2
2
4
f (x)
f ′(x)
x
y
Figure 3.5: A plot of f (x) = x4/4 + x3/3 − x2 andf ′(x) = x3
+ x2 − 2x.
Hence we have seen that if f ′(x) is zero and increasing at a point, then f (x) has a
local minimum at the point. If f ′(x) is zero and decreasing at a point then f (x) has
a local maximum at the point. Thus, we see that we can gain information about
f (x) by studying how f ′(x) changes. This leads us to our next section.
calculus 59
Exercises for Section 3.2
In the following exercises, find all critical points and identify them as local maximum
points, local minimum points, or neither.
(1) y = x2 − x ➠
(2) y = 2 + 3x − x3➠
(3) y = x3 − 9x2+ 24x ➠
(4) y = x4 − 2x2+ 3 ➠
(5) y = 3x4 − 4x3➠
(6) y = (x2 − 1)/x ➠
(7) f (x) = |x2 − 121| ➠
(8) Let f (x) = ax2+ bx + c with a , 0. Show that f (x) has exactly one critical point
using the first derivative test. Give conditions on a and b which guarantee that
the critical point will be a maximum. ➠
60
3.3 Concavity and Inflection Points
We know that the sign of the derivative tells us whether a function is increasing or
decreasing. Likewise, the sign of the second derivative f ′′(x) tells us whether f ′(x)
is increasing or decreasing.
Suppose that f ′′(a) > 0. This means that near x = a, f ′(x) is increasing. If
f ′(a) > 0, this means that f (x) slopes up and is getting steeper. If f ′(a) < 0, this
means that f (x) slopes down and is getting less steep. These two situations are
shown in Figure 3.6. A curve that is shaped like this is called concave up.
Concave Up Concave Up
Here f ′(x) > 0 and f ′′(x) > 0. This meansthat f (x) slopes up and is getting steeper.
Here f ′(x) < 0 and f ′′(x) > 0. This meansthat f (x) slopes down and is getting less
steep.
Figure 3.6: Examples of when a curve is concave up.
Now suppose that f ′′(a) < 0. This means that near x = a, f ′(x) is decreasing. If
f ′(a) > 0, this means that f (x) slopes up and is getting less steep; if f ′(a) < 0, this
means that f (x) slopes down and is getting steeper. These two situations are shown
in Figure 3.7. A curve that is shaped like this is called concave down.
Concave Down Concave Down
Here f ′(x) < 0 and f ′′(x) < 0. Thismeans that f (x) slopes down and is gettingsteeper.
Here f ′(x) > 0 and f ′′(x) < 0. This meansthat f (x) slopes up and is getting lesssteep.
Figure 3.7: Examples of when a curve is concavedown.
If we are trying to understand the shape of the graph of a function, knowing
where it is concave up and concave down helps us to get a more accurate picture. It
calculus 61
is worth summarizing what we have seen already in to a single theorem.
Theorem 3.3.1 (Test for Concavity) Suppose that f ′′(x) exists on an inter-
val.
(a) If f ′′(x) > 0 on an interval, then f (x) is concave up on that interval.
(b) If f ′′(x) < 0 on an interval, then f (x) is concave down on that interval.
Of particular interest are points at which the concavity changes from up to down
or down to up.
Definition If f (x) is continuous and its concavity changes either from up to
down or down to up at x = a, then f (x) has an inflection point at x = a.
It is instructive to see some examples and nonexamples of inflection points.
This is an inflection point. The concav-ity changes from concave up to concavedown.
This is not an inflection point. The curveis concave down on either side of the point.
This is an inflection point. The concav-ity changes from concave up to concavedown.
This is not an inflection point. The curveis concave down on either side of the point.
We identify inflection points by first finding where f ′′(x) is zero and then checking
to see whether f ′′(x) does in fact go from positive to negative or negative to positive
at these points.
Warning Even if f ′′(a) = 0, the point determined by x = a might not be an
inflection point.
62
Example 3.3.2 Describe the concavity of f (x) = x3 − x.
Solution To start, compute the first and second derivative of f (x) with respect
to x,
f ′(x) = 3x2 − 1 and f ′′(x) = 6x.
Since f ′′(0) = 0, there is potentially an inflection point at zero. Since f ′′(x) > 0
when x > 0 and f ′′(x) < 0 when x < 0 the concavity does change from down to
up at zero—there is an inflection point at x = 0. The curve is concave down for
all x < 0 and concave up for all x > 0, see Figure 3.8.
−1.5 −1 −0.5 0.5 1 1.5
−2
2
f (x)
f ′′(x)
x
y
Figure 3.8: A plot of f (x) = x3 − x and f ′′(x) = 6x.We can see that the concavity change at x = 0.
Note that we need to compute and analyze the second derivative to understand
concavity, so we may as well try to use the second derivative test for maxima and
minima. If for some reason this fails we can then try one of the other tests.
calculus 63
Exercises for Section 3.3
In the following exercises, describe the concavity of the functions.
(1) y = x2 − x ➠
(2) y = 2 + 3x − x3➠
(3) y = x3 − 9x2+ 24x ➠
(4) y = x4 − 2x2+ 3 ➠
(5) y = 3x4 − 4x3➠
(6) y = (x2 − 1)/x ➠
(7) y = 3x2 −1
x2 ➠
(8) y = x5 − x ➠
(9) y = x + 1/x ➠
(10) y = x2+ 1/x ➠
(11) Identify the intervals on which the graph of the function f (x) = x4 − 4x3+ 10
is of one of these four shapes: concave up and increasing; concave up and
decreasing; concave down and increasing; concave down and decreasing. ➠
64
3.4 The Second Derivative Test
Recall the first derivative test, Theorem 3.2.1:
• If f ′(x) > 0 to the left of a and f ′(x) < 0 to the right of a, then f (a) is a local
maximum.
• If f ′(x) < 0 to the left of a and f ′(x) > 0 to the right of a, then f (a) is a local
minimum.
If f ′(x) changes from positive to negative it is decreasing. In this case, f ′′(x)
might be negative, and if in fact f ′′(x) is negative then f ′(x) is definitely decreasing,
so there is a local maximum at the point in question. On the other hand, if f ′(x)
changes from negative to positive it is increasing. Again, this means that f ′′(x)
might be positive, and if in fact f ′′(x) is positive then f ′(x) is definitely increasing,
so there is a local minimum at the point in question. We summarize this as the
second derivative test.
Theorem 3.4.1 (Second Derivative Test) Suppose that f ′′(x) is continu-
ous on an open interval and that f ′(a) = 0 for some value of a in that
interval.
• If f ′′(a) < 0, then f (x) has a local maximum at a.
• If f ′′(a) > 0, then f (x) has a local minimum at a.
• If f ′′(a) = 0, then the test is inconclusive. In this case, f (x) may or may
not have a local extremum at x = a.
The second derivative test is often the easiest way to identify local maximum and
minimum points. Sometimes the test fails and sometimes the second derivative is
quite difficult to evaluate. In such cases we must fall back on one of the previous
tests.
calculus 65
Example 3.4.2 Once again, consider the function
f (x) =x4
4+
x3
3− x2
Use the second derivative test, Theorem 3.4.1, to locate the local extrema of
f (x).
Solution Start by computing
f ′(x) = x3+ x2 − 2x and f ′′(x) = 3x2
+ 2x − 2.
Using the same technique as used in the solution of Example 3.2.2, we find that
f ′(−2) = 0, f ′(0) = 0, f ′(1) = 0.
Now we’ll attempt to use the second derivative test, Theorem 3.4.1,
f ′′(−2) = 6, f ′′(0) = −2, f ′′(1) = 3.
Hence we see that f (x) has a local minimum at x = −2, a local maximum at
x = 0, and a local minimum at x = 1, see Figure 3.9.
−3 −2 −1 1 2
−4
−2
2
4
6
f (x)
f ′′(x)
x
y
Figure 3.9: A plot of f (x) = x4/4 + x3/3 − x2 andf ′′(x) = 3x2
+ 2x − 2.
Warning If f ′′(a) = 0, then the second derivative test gives no information on
whether x = a is a local extremum.
66
Exercises for Section 3.4
Find all local maximum and minimum points by the second derivative test.
(1) y = x2 − x ➠
(2) y = 2 + 3x − x3➠
(3) y = x3 − 9x2+ 24x ➠
(4) y = x4 − 2x2+ 3 ➠
(5) y = 3x4 − 4x3➠
(6) y = (x2 − 1)/x ➠
(7) y = 3x2 −1
x2 ➠
(8) y = x5 − x ➠
(9) y = x + 1/x ➠
(10) y = x2+ 1/x ➠
calculus 67
3.5 Sketching the Plot of a Function
In this section, we will give some general guidelines for sketching the plot of a
function.
Procedure for Sketching the Plots of Functions
• Find the y-intercept, this is the point (0, f (0)). Place this point on your
graph.
• Find candidates for vertical asymptotes, these are points where f (x) is
undefined.
• Compute f ′(x) and f ′′(x).
• Find the critical points, the points where f ′(x) = 0.
• Use the second derivative test to identify local extrema and/or find the
intervals where your function is increasing/decreasing.
• Find the candidates for inflection points, the points where f ′′(x) = 0.
• Identify inflection points and concavity.
• If possible find the x-intercepts, the points where f (x) = 0. Place these
points on your graph.
• Find horizontal asymptotes.
• Determine an interval that shows all relevant behavior.
At this point you should be able to sketch the plot of your function.
Let’s see this procedure in action. We’ll sketch the plot of 2x3 − 3x2 − 12x.
Following our guidelines above, we start by computing f (0) = 0. Hence we see that
the y-intercept is (0, 0). Place this point on your plot, see Figure 3.10.
−2 −1 1 2 3 4
−20
−10
10
20
x
y
Figure 3.10: We start by placing the point (0, 0).
68
Note that there are no vertical asymptotes as our function is defined for all real
numbers. Now compute f ′(x) and f ′′(x),
f ′(x) = 6x2 − 6x − 12 and f ′′(x) = 12x − 6.
The critical points are where f ′(x) = 0, thus we need to solve 6x2 − 6x − 12 = 0
for x. Write
6x2 − 6x − 12 = 0
x2 − x − 2 = 0
(x − 2)(x + 1) = 0.
Thus
f ′(2) = 0 and f ′(−1) = 0.
Mark the critical points x = 2 and x = −1 on your plot, see Figure 3.11.
−2 −1 1 2 3 4
−20
−10
10
20
x
y
Figure 3.11: Now we add the critical points x = −1and x = 2.
Check the second derivative evaluated at the critical points. In this case,
f ′′(−1) = −18 and f ′′(2) = 18,
hence x = −1, corresponding to the point (−1, 7) is a local maximum and x = 2,
corresponding to the point (2,−20) is local minimum of f (x). Moreover, this tells us
that our function is increasing on [−2,−1), decreasing on (−1, 2), and increasing on
(2, 4]. Identify this on your plot, see Figure 3.12.−2 −1 1 2 3 4
−20
−10
10
20
x
y
Figure 3.12: We have identified the local extremaof f (x) and where this function is increasing anddecreasing.
The candidates for the inflection points are where f ′′(x) = 0, thus we need to
solve 12x − 6 = 0 for x. Write
12x − 6 = 0
x − 1/2 = 0
x = 1/2.
Thus f ′′(1/2) = 0. Checking points, f ′′(0) = −6 and f ′′(1) = 6. Hence x = 1/2 is an
inflection point, with f (x) concave down to the left of x = 1/2 and f (x) concave up
to the right of x = 1/2. We can add this information to our plot, see Figure 3.13.
calculus 69
Finally, in this case, f (x) = 2x3 − 3x2 − 12x, we can find the x-intercepts. Write
2x3 − 3x2 − 12x = 0
x(2x2 − 3x − 12) = 0.
Using the quadratic formula, we see that the x-intercepts of f (x) are
x = 0, x =3 −√
105
4, x =
3 +√
105
4.
Since all of this behavior as described above occurs on the interval [−2, 4], we now
have a complete sketch of f (x) on this interval, see the figure below. −2 −1 1 2 3 4
−20
−10
10
20
x
y
Figure 3.13: We identify the inflection point and notethat the curve is concave down when x < 1/2 andconcave up when x > 1/2.
−2 −1 1 2 3 4
−20
−10
10
20
x
y
70
Exercises for Section 3.5
Sketch the curves via the procedure outlined in this section. Clearly identify any
interesting features, including local maximum and minimum points, inflection
points, asymptotes, and intercepts.
(1) y = x5 − x ➠
(2) y = x(x2+ 1) ➠
(3) y = 2√
x − x ➠
(4) y = x3+ 6x2
+ 9x ➠
(5) y = x3 − 3x2 − 9x + 5 ➠
(6) y = x5 − 5x4+ 5x3
➠
(7) y = x + 1/x ➠
(8) y = x2+ 1/x ➠
4 The Product Rule and Quotient Rule
4.1 The Product Rule
Consider the product of two simple functions, say
f (x) · g(x)
where f (x) = x2+ 1 and g(x) = x3 − 3x. An obvious guess for the derivative of
f (x)g(x) is the product of the derivatives:
f ′(x)g′(x) = (2x)(3x2 − 3)
= 6x3 − 6x.
Is this guess correct? We can check by rewriting f (x) and g(x) and doing the
calculation in a way that is known to work. Write
f (x)g(x) = (x2+ 1)(x3 − 3x)
= x5 − 3x3+ x3 − 3x
= x5 − 2x3 − 3x.
Henced
dxf (x)g(x) = 5x4 − 6x2 − 3,
so we see thatd
dxf (x)g(x) , f ′(x)g′(x).
So the derivative of f (x)g(x) is not as simple as f ′(x)g′(x). Never fear, we have a
rule for exactly this situation.
72
Theorem 4.1.1 (The Product Rule) If f (x) and g(x) are differentiable, then
d
dxf (x)g(x) = f (x)g′(x) + f ′(x)g(x).
f (x)
g(x)
f (x)g(x)
f (a)g′(a)h + f ′(a)hg(a)
f ′(a)h
g′(a)h
︸ ︷︷ ︸
h
ax
y
Figure 4.1: A geometric interpretation of the productrule. Since every point on f (x)g(x) is the product ofthe corresponding points on f (x) and g(x), increasinga by a “small amount” h, increases f (a)g(a) by thesum of f (a)g′(a)h and f ′(a)hg(a). Hence,
∆y
∆x≈ f (a)g′(a)h + f ′(a)g(a)h
h= f (a)g′(a)+f ′(a)g(a).
Proof From the limit definition of the derivative, write
d
dx(f (x)g(x)) = lim
h→0
f (x + h)g(x + h) − f (x)g(x)
h
Now we use the exact same trick we used in the proof of Theorem 1.2.2, we add
0 = −f (x + h)g(x) + f (x + h)g(x):
= limh→0
f (x + h)g(x + h)−f (x + h)g(x) + f (x + h)g(x) − f (x)g(x)
h
= limh→0
f (x + h)g(x + h) − f (x + h)g(x)
h+ lim
h→0
f (x + h)g(x) − f (x)g(x)
h.
Now since both f (x) and g(x) are differentiable, they are continuous, see Theo-
rem 2.1.3. Hence
= limh→0
f (x + h)g(x + h) − g(x)
h+ lim
h→0
f (x + h) − f (x)
hg(x)
= limh→0
f (x + h) limh→0
g(x + h) − g(x)
h+ lim
h→0
f (x + h) − f (x)
hlimh→0
g(x)
= f (x)g′(x) + f ′(x)g(x).
Let’s return to the example with which we started.
Example 4.1.2 Let f (x) = (x2+ 1) and g(x) = (x3 − 3x). Compute:
d
dxf (x)g(x).
calculus 73
Solution Write
d
dxf (x)g(x) = f (x)g′(x) + f ′(x)g(x)
= (x2+ 1)(3x2 − 3) + 2x(x3 − 3x).
We could stop here—but we should show that expanding this out recovers our
previous result. Write
(x2+ 1)(3x2 − 3) + 2x(x3 − 3x) = 3x4 − 3x2
+ 3x2 − 3 + 2x4 − 6x2
= 5x4 − 6x2 − 3,
which is precisely what we obtained before.
74
Exercises for Section 4.1
Compute:
(1)d
dxx3(x3 − 5x + 10) ➠
(2)d
dx(x2+5x−3)(x5−6x3
+3x2−7x+1)
➠
(3)d
dxe2x=
d
dx(ex · ex ) ➠
(4)d
dxe3x
➠
(5)d
dx3x2e4x
➠
(6)d
dx
3ex
x16 ➠
(7) Use the product rule to compute the derivative of f (x) = (2x − 3)2 with respect
to x. Sketch the function. Find an equation of the tangent line to the curve at
x = 2. Sketch the tangent line at x = 2. ➠
Use the following table to compute solve the next 4 problems. Noted
dxf (x)
∣∣∣∣∣x=a
is
the derivative of f (x) evaluated at x = a.
x 1 2 3 4
f (x) −2 −3 1 4
f ′(x) −1 0 3 5
g(x) 1 4 2 −1
g′(x) 2 −1 −2 −3
(8)d
dxf (x)g(x)
∣∣∣∣∣x=2
➠
(9)d
dxxf (x)
∣∣∣∣∣x=3
➠
(10)d
dxxg(x)
∣∣∣∣∣x=4
➠
(11)d
dxf (x)g(x)
∣∣∣∣∣x=1
➠
(12) Suppose that f (x), g(x), and h(x) are differentiable functions. Show that
d
dxf (x) · g(x) · h(x) = f (x)g(x)h′(x) + f (x)g′(x)h(x) + f ′(x)g(x)h(x).
➠
calculus 75
4.2 The Quotient Rule
We’d like to have a formula to compute
d
dx
f (x)
g(x)
if we already know f ′(x) and g′(x). Instead of attacking this problem head-on, let’s
notice that we’ve already done part of the problem: f (x)/g(x) = f (x) · (1/g(x)), that
is, this is really a product, and we can compute the derivative if we know f ′(x) and
(1/g(x))′. This brings us to our next derivative rule.
Theorem 4.2.1 (The Quotient Rule) If f (x) and g(x) are differentiable, then
d
dx
f (x)
g(x)=
f ′(x)g(x) − f (x)g′(x)
g(x)2.
Proof First note that if we knew how to compute
d
dx
1
g(x)
then we could use the product rule to complete our proof. Write
d
dx
1
g(x)= lim
h→0
1g(x+h)
− 1g(x)
h
= limh→0
g(x)−g(x+h)
g(x+h)g(x)
h
= limh→0
g(x) − g(x + h)
g(x + h)g(x)h
= limh→0−g(x + h) − g(x)
h
1
g(x + h)g(x)
= −g′(x)
g(x)2.
76
Now we can put this together with the product rule:
d
dx
f (x)
g(x)= f (x)
−g′(x)
g(x)2+ f ′(x)
1
g(x)
=−f (x)g′(x) + f ′(x)g(x)
g(x)2
=f ′(x)g(x) − f (x)g′(x)
g(x)2.
Example 4.2.2 Compute:d
dx
x2+ 1
x3 − 3x.
Solution Write
d
dx
x2+ 1
x3 − 3x=
2x(x3 − 3x) − (x2+ 1)(3x2 − 3)
(x3 − 3x)2
=−x4 − 6x2
+ 3
(x3 − 3x)2.
It is often possible to calculate derivatives in more than one way, as we have
already seen. Since every quotient can be written as a product, it is always possible
to use the product rule to compute the derivative, though it is not always simpler.
Example 4.2.3 Computed
dx
625 − x2
√x
in two ways. First using the quotient rule and then using the product rule.
Solution First, we’ll compute the derivative using the quotient rule. Write
d
dx
625 − x2
√x
=
(−2x)(√
x)
− (625 − x2)(
12 x−1/2
)
x.
calculus 77
Second, we’ll compute the derivative using the product rule:
d
dx
625 − x2
√x
=d
dx
(
625 − x2)
x−1/2
=
(
625 − x2)(
−x−3/2
2
)
+ (−2x)(
x−1/2)
.
With a bit of algebra, both of these simplify to
−3x2+ 625
2x3/2.
78
Exercises for Section 4.2
Find the derivatives of the following functions using the quotient rule.
(1)x3
x3 − 5x + 10 ➠
(2)x2+ 5x − 3
x5 − 6x3 + 3x2 − 7x + 1 ➠
(3)ex − 4
2x ➠
(4)2 − x −
√x
x + 2 ➠
(5) Find an equation for the tangent line to f (x) = (x2 − 4)/(5 − x) at x = 3. ➠
(6) Find an equation for the tangent line to f (x) = (x − 2)/(x3+ 4x − 1) at x = 1.
➠
(7) The curve y = 1/(1 + x2) is an example of a class of curves each of which is
called a witch of Agnesi. Find the tangent line to the curve at x = 5. Note, the
word witch here is due to a mistranslation. ➠
Use the following table to compute solve the next 4 problems. Noted
dxf (x)
∣∣∣∣∣x=a
is
the derivative of f (x) evaluated at x = a.
x 1 2 3 4
f (x) −2 −3 1 4
f ′(x) −1 0 3 5
g(x) 1 4 2 −1
g′(x) 2 −1 −2 −3
(8)d
dx
f (x)
g(x)
∣∣∣∣∣x=2
➠
(9)d
dx
f (x)
x
∣∣∣∣∣x=3
➠
(10)d
dx
xf (x)
g(x)
∣∣∣∣∣x=4
➠
(11)d
dx
f (x)g(x)
x
∣∣∣∣∣x=1
➠
(12) If f ′(4) = 5, g′(4) = 12, f (4)g(4) = 2, and g(4) = 6, compute f (4) andd
dx
f (x)
g(x)at
4. ➠
5 The Chain Rule
So far we have seen how to compute the derivative of a function built up from other
functions by addition, subtraction, multiplication and division. There is another
very important way that we combine functions: composition. The chain rule allows
us to deal with this case.
5.1 The Chain Rule
Consider
h(x) = (1 + 2x)5.
While there are several different ways to differentiate this function, if we let
f (x) = x5 and g(x) = 1 + 2x, then we can express h(x) = f (g(x)). The question is,
can we compute the derivative of a composition of functions using the derivatives of
the constituents f (x) and g(x)? To do so, we need the chain rule.
f ′(g(a))g′(a)h
g′(a)h
h︷
︸︸
︷
g(x)
y
x
a
Figure 5.1: A geometric interpretation of the chainrule. Increasing a by a “small amount” h, increasesf (g(a)) by f ′(g(a))g′(a)h. Hence,
∆y
∆x≈ f ′(g(a))g′(a)h
h= f ′(g(a))g′(a).
Theorem 5.1.1 (Chain Rule) If f (x) and g(x) are differentiable, then
d
dxf (g(x)) = f ′(g(x))g′(x).
Proof Let g0 some x-value and consider the following:
f ′(g0) = limh→0
f (g0 + h) − f (g0)
h.
80
Set h = g − g0 and we have
f ′(g0) = limg→g0
f (g) − f (g0)
g − g0. (5.1)
At this point, we might like to set g = g(x +h) and g0 = g(x); however, we cannot
as we cannot be sure that
g(x + h) − g(x) , 0 when h , 0.
To overcome this difficulty, let E(g) be the difference between f (g) and the tangent
line of f (x) at x = g, see the figure below:
︸ ︷︷ ︸
g−g0
f ′(g0)(g − g0)
E(g)(g − g0)
g0 g
f (g0)
f (g)
x
y
Hence we see that
f (g) − f (g0) =(
f ′(g0) + E(g))
(g − g0), (5.2)
and sof (g) − f (g0)
g − g0= f ′(g0) + E(g).
calculus 81
Combining this with Equation 5.1, we have that
f ′(g0) = limg→g0
f (g) − f (g0)
g − g0
= limg→g0
f ′(g0) + E(g)
= f ′(g0) + limg→g0
E(g),
and hence it follows that limg→g0
E(g) = 0. At this point, we may return to the
“well-worn path.” Starting with Equation 5.2, divide both sides by h and set
g = g(x + h) and g0 = g(x)
f (g(x + h)) − f (g(x))
h=
(
f ′(g(x)) + E(g(x))) g(x + h) − g(x)
h.
Taking the limit as h approaches 0, we see
limh→0
f (g(x + h)) − f (g(x))
h= lim
h→0
(
f ′(g(x)) + E(g(x))) g(x + h) − g(x)
h
= limh→0
(
f ′(g(x)) + E(g(x)))
limh→0
g(x + h) − g(x)
h
= f ′(g(x))g′(x).
Hence,d
dxf (g(x)) = f ′(g(x))g′(x).
It will take a bit of practice to make the use of the chain rule come naturally—it
is more complicated than the earlier differentiation rules we have seen. Let’s return
to our motivating example.
Example 5.1.2 Compute:d
dx(1 + 2x)5
Solution Set f (x) = x5 and g(x) = 1 + 2x, now
f ′(x) = 5x4 and g′(x) = 2.
82
Hence
d
dx(1 + 2x)5
=d
dxf (g(x))
= f ′(g(x))g′(x)
= 5(1 + 2x)4 · 2
= 10(1 + 2x)4.
Let’s see a more complicated chain of compositions.
Example 5.1.3 Compute:d
dx
√
1 +√
x
Solution Set f (x) =√
x and g(x) = 1 + x. Hence,
√
1 +√
x = f (g(f (x))) andd
dxf (g(f (x))) = f ′(g(f (x)))g′(f (x))f ′(x).
Since
f ′(x) =1
2√
xand g′(x) = 1
We have thatd
dx
√
1 +√
x =1
2√
1 +√
x
· 1 ·1
2√
x.
Using the chain rule, the power rule, and the product rule it is possible to avoid
using the quotient rule entirely.
Example 5.1.4 Compute:d
dx
x3
x2 + 1
Solution Rewriting this as
d
dxx3(x2
+ 1)−1,
calculus 83
set f (x) = x−1 and g(x) = x2+ 1. Now
x3(x2+1)−1
= x3f (g(x)) andd
dxx3f (g(x)) = 3x2f (g(x))+x3f ′(g(x))g′(x).
Since f ′(x) =−1
x2and g′(x) = 2x, write
d
dx
x3
x2 + 1=
3x2
x2 + 1−
2x4
(x2 + 1)2.
84
Exercises for Section 5.1
Compute the derivatives of the functions. For extra practice, and to check your
answers, do some of these in more than one way if possible.
(1) x4 − 3x3+ (1/2)x2
+ 7x − π ➠
(2) x3 − 2x2+ 4√
x ➠
(3) (x2+ 1)3
➠
(4) x√
169 − x2➠
(5) (x2 − 4x + 5)√
25 − x2➠
(6)√
r2 − x2, r is a constant ➠
(7)√
1 + x4➠
(8)1
√
5 −√
x
. ➠
(9) (1 + 3x)2➠
(10)(x2+ x + 1)
(1 − x) ➠
(11)
√25 − x2
x ➠
(12)
√
169
x− x ➠
(13)√
x3 − x2 − (1/x) ➠
(14) 100/(100 − x2)3/2➠
(15)3√x + x3
➠
(16)
√
(x2 + 1)2 +
√
1 + (x2 + 1)2➠
(17) (x + 8)5➠
(18) (4 − x)3➠
(19) (x2+ 5)3
➠
(20) (6 − 2x2)3➠
(21) (1 − 4x3)−2➠
(22) 5(x + 1 − 1/x) ➠
(23) 4(2x2 − x + 3)−2➠
(24)1
1 + 1/x ➠
(25)−3
4x2 − 2x + 1 ➠
(26) (x2+ 1)(5 − 2x)/2 ➠
(27) (3x2+ 1)(2x − 4)3
➠
(28)x + 1
x − 1 ➠
(29)x2 − 1
x2 + 1 ➠
(30)(x − 1)(x − 2)
x − 3 ➠
(31)2x−1 − x−2
3x−1 − 4x−2 ➠
(32) 3(x2+ 1)(2x2 − 1)(2x + 3) ➠
(33)1
(2x + 1)(x − 3) ➠
(34) ((2x + 1)−1+ 3)−1
➠
(35) (2x + 1)3(x2+ 1)2
➠
calculus 85
(36) Find an equation for the tangent line to f (x) = (x − 2)1/3/(x3+ 4x − 1)2 at x = 1.
➠
(37) Find an equation for the tangent line to y = 9x−2 at (3, 1). ➠
(38) Find an equation for the tangent line to (x2 − 4x + 5)√
25 − x2 at (3, 8). ➠
(39) Find an equation for the tangent line to(x2+ x + 1)
(1 − x)at (2,−7). ➠
(40) Find an equation for the tangent line to
√
(x2 + 1)2 +
√
1 + (x2 + 1)2 at (1,
√
4 +√
5).
➠
86
5.2 Implicit Differentiation
The functions we’ve been dealing with so far have been explicit functions, meaning
that the dependent variable is written in terms of the independent variable. For
example:
y = 3x2 − 2x + 1, y = e3x , y =x − 2
x2 − 3x + 2.
However, there are another type of functions, called implicit functions. In this case,
the dependent variable is not stated explicitly in terms of the independent variable.
For example:
x2+ y2
= 4, x3+ y3
= 9xy, x4+ 3x2
= x2/3+ y2/3
= 1.
Your inclination might be simply to solve each of these for y and go merrily on your
way. However this can be difficult and it may require two branches, for example to
explicitly plot x2+ y2
= 4, one needs both y =√
4 − x2 and y = −√
4 − x2. Moreover,
it not even be possible to solve for y. To deal with such situations, we use implicit
differentiation. Let’s see an illustrative example:
Example 5.2.1 Consider the curve defined by
x3+ y3
= 9xy.
(a) Computedy
dx.
(b) Find the slope of the tangent line at (4, 2).−6 −4 −2 2 4 6
−6
−4
−2
2
4
6
x
y
Figure 5.2: A plot of x3+ y3
= 9xy. While this isnot a function of y in terms of x, the equation stilldefines a relation between x and y.
Solution Starting with
x3+ y3
= 9xy,
we apply the differential operatord
dxto both sides of the equation to obtain
d
dx
(
x3+ y3
)
=d
dx9xy.
Applying the sum rule we see
d
dxx3+
d
dxy3=
d
dx9xy.
calculus 87
Let’s examine each of these terms in turn. To start
d
dxx3= 3x2.
On the other handd
dxy3 is somewhat different. Here you imagine that y = y(x),
and hence by the chain rule
d
dxy3=
d
dx(y(x))3
= 3(y(x))2 · y′(x)
= 3y2 dy
dx.
Considering the final termd
dx9xy, we again imagine that y = y(x). Hence
d
dx9xy = 9
d
dxx · y(x)
= 9(
x · y′(x) + y(x))
= 9xdy
dx+ 9y.
Putting this all together we are left with the equation
3x2+ 3y2 dy
dx= 9x
dy
dx+ 9y.
At this point, we solve fordy
dx. Write
3x2+ 3y2 dy
dx= 9x
dy
dx+ 9y
3y2 dy
dx− 9x
dy
dx= 9y − 3x2
dy
dx
(
3y2 − 9x)
= 9y − 3x2
dy
dx=
9y − 3x2
3y2 − 9x=
3y − x2
y2 − 3x.
For the second part of the problem, we simply plug x = 4 and y = 2 into the
formula above, hence the slope of the tangent line at (4, 2) is5
4, see Figure 5.3.
−6 −4 −2 2 4 6
−6
−4
−2
2
4
6
x
y
Figure 5.3: A plot of x3+ y3
= 9xy along with thetangent line at (4, 2).
88
You might think that the step in which we solve fordy
dxcould sometimes be
difficult—after all, we’re using implicit differentiation here instead of the more difficult
task of solving the equation x3+ y3
= 9xy for y, so maybe there are functions where
after taking the derivative we obtain something where it is hard to solve fordy
dx. In
fact, this never happens. All occurrencesdy
dxarise from applying the chain rule, and
whenever the chain rule is used it deposits a singledy
dxmultiplied by some other
expression. Hence our expression in linear indy
dx, it will always be possible to group
the terms containingdy
dxtogether and factor out the
dy
dx, just as in the previous
example.
The Derivative of Inverse Functions
Geometrically, there is a close relationship between the plots of ex and ln(x), they
are reflections of each other over the line y = x, see Figure 5.4. One may suspect
that we can use the fact thatd
dxex= ex , to deduce the derivative of ln(x). We will
use implicit differentiation to exploit this relationship computationally.−6 −4 −2 2 4 6
−6
−4
−2
2
4
6
ex
ln(x)
x
y
Figure 5.4: A plot of ex and ln(x). Since they areinverse functions, they are reflections of each otheracross the line y = x.
Theorem 5.2.2 (The Derivative of the Natural Logrithm)
d
dxln(x) =
1
x.
Proof Recall
ln(x) = y ⇔ ey= x.
calculus 89
Hence
ey= x
d
dxey=
d
dxx Differentiate both sides.
ey dy
dx= 1 Implicit differentiation.
dy
dx=
1
ey=
1
x.
Since y = ln(x),d
dxln(x) =
1
x.
There is one catch to the proof given above. To writed
dx(ey) = ey dy
dxwe need
to know that the function y has a derivative. All we have shown is that if it
has a derivative then that derivative must be 1/x. The Inverse Function Theorem
guarantees this.
Theorem 5.2.3 (Inverse Function Theorem) If f (x) is a differentiable func-
tion, and f ′(x) is continuous, and f ′(a) , 0, then
(a) f −1(y) is defined for y near f (a),
(b) f −1(y) is differentiable near f (a),
(c)d
dyf −1(y) is continuous near f (a), and
(d)d
dyf −1(y) =
1
f ′(f −1(y)).
90
Exercises for Section 5.2
Computedy
dx:
(1) x2+ y2
= 4 ➠
(2) y2= 1 + x2
➠
(3) x2+ xy + y2
= 7 ➠
(4) x3+ xy2
= y3+ yx2
➠
(5) x2y − y3= 6 ➠
(6)√
x +√
y = 9 ➠
(7) xy3/2+ 4 = 2x + y ➠
(8)1
x+
1
y= 7 ➠
(9) A hyperbola passing through (8, 6) consists of all points whose distance from
the origin is a constant more than its distance from the point (5,2). Find the
slope of the tangent line to the hyperbola at (8, 6). ➠
(10) The graph of the equation x2 − xy + y2= 9 is an ellipse. Find the lines tangent
to this curve at the two points where it intersects the x-axis. Show that these
lines are parallel. ➠
(11) Repeat the previous problem for the points at which the ellipse intersects the
y-axis. ➠
(12) Find the points on the ellipse from the previous two problems where the slope
is horizontal and where it is vertical. ➠
(13) Find an equation for the tangent line to x4= y2
+ x2 at (2,√
12). This curve is
the kampyle of Eudoxus. ➠
(14) Find an equation for the tangent line to x2/3+ y2/3
= a2/3 at a point (x1, y1) on
the curve, with x1 , 0 and y1 , 0. This curve is an astroid. ➠
(15) Find an equation for the tangent line to (x2+ y2)2
= x2 − y2 at a point (x1, y1)
on the curve, with x1 , 0,−1, 1. This curve is a lemniscate. ➠
calculus 91
5.3 Logarithmic Differentiation
Logarithms were originally developed as a computational tool. The key fact that
made this possible is that
logb(xy) = logb(x) + logb(y).
While this may seem quite abstract, before the days of calculators and computers, 1 2 3 4 5 6 7
−4
−2
2
x
y
Figure 5.5: A plot of ln(x). Here we see that
ln(2 · 3) = ln(2) + ln(3).
this was critical knowledge for anyone in a computational discipline. Suppose you
wanted to compute
138 · 23.4
You would start by writing both in scientific notation
(
1.38 · 102)
·(
2.34 · 101)
.
Next you would use a log-table, which gives log10(N) for values of N ranging between
0 and 9. We’ve reproduced part of such a table below.
N 0 1 2 3 4 5 6 7 8 9
1.3 0.1139 0.1173 0.1206 0.1239 0.1271 0.1303 0.1335 0.1367 0.1399 0.1430
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 0.3617 0.3636 0.3655 0.3674 0.3692 0.3711 0.3729 0.3747 0.3766 0.3784
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 0.5052 0.5065 0.5079 0.5092 0.5105 0.5119 0.5132 0.5145 0.5159 0.5172
Figure 5.6: Part of a base-10 logarithm table.
From the table, we see that
log10(1.38) ≈ 0.1399 and log10(2.34) ≈ 0.3692
Add these number together to get 0.5091. Essentially, we know the following at this
point:log10(?) = log10(1.38) + log10(2.34)
≈ ≈ ≈
0.5091 = 0.1399 + 0.3692
92
Using the table again, we see that log10(3.23) ≈ 0.5091. Since we were working
in scientific notation, we need to multiply this by 103. Our final answer is
3230 ≈ 138 · 23.4
Since 138 · 23.4 = 3229.2, this is a good approximation. The moral is:
Logarithms allow us to use addition in place of of multiplication.
When taking derivatives, both the product rule and the quotient rule can be
cumbersome to use. Logarithms will save the day. A key point is the following
d
dxln(f (x)) =
1
f (x)· f ′(x) =
f ′(x)
f (x)
which follows from the chain rule. Let’s look at an illustrative example to see how
this is actually used.
Example 5.3.1 Compute:d
dx
x9e4x
√x2 + 4
Recall the properties of logarithms:
• logb(xy) = logb(x) + logb(y)
• logb(x/y) = logb(x) − logb(y)
• logb(xy) = y logb(x)
Solution While we could use the product and quotient rule to solve this prob-
lem, it would be tedious. Start by taking the logarithm of the function to be
differentiated.
ln
(
x9e4x
√x2 + 4
)
= ln(
x9e4x)
− ln(√
x2 + 4)
= ln(
x9)
+ ln(
e4x)
− ln(
(x2+ 4)1/2
)
= 9 ln(x) + 4x −1
2ln(x2
+ 4).
Setting f (x) =x9e4x
√x2 + 4
, we can write
ln(f (x)) = 9 ln(x) + 4x −1
2ln(x2
+ 4).
calculus 93
Differentiating both sides, we find
f ′(x)
f (x)=
9
x+ 4 − x
x2 + 4.
Finally we solve for f ′(x), write
f ′(x) =
(9
x+ 4 −
x
x2 + 4
) (
x9e4x
√x2 + 4
)
.
The process above is called logarithmic differentiation. Logarithmic differentiation
allows us to compute new derivatives too.
Example 5.3.2 Compute:d
dxxx
Solution The function xx is tricky to differentiate. We cannot use the power
rule, as the exponent is not a constant. However, if we set f (x) = xx we can
write
ln(f (x)) = ln (xx )
= x ln(x).
Differentiating both sides, we find
f ′(x)
f (x)= x ·
1
x+ ln(x)
= 1 + ln(x).
Now we can solve for f ′(x),
f ′(x) = xx+ xx ln(x).
Finally recall that previously we only proved the power rule, Theorem 2.2.2, for
positive exponents. Now we’ll use logarithmic differentiation to give a proof for all
real-valued exponents. We restate the power rule for convenience sake:
94
Theorem 5.3.3 (Power Rule) For any real number n,
d
dxxn= nxn−1.
Proof We will use logarithmic differentiation. Set f (x) = xn. Write
ln(f (x)) = ln (xn)
= n ln(x).
Now differentiate both sides, and solve for f ′(x)
f ′(x)
f (x)=
n
x
f ′(x) =nf (x)
x
= nxn−1.
Thus we see that the power rule holds for all real-valued exponents.
While logarithmic differentiation might seem strange and new at first, with a little
practice it will seem much more natural to you.
calculus 95
Exercises for Section 5.3
Use logarithmic differentiation to compute the following:
(1)d
dx(x + 1)3
√x4 + 5 ➠
(2)d
dxx2e5x
➠
(3)d
dxx ln(x)
➠
(4)d
dxx100x
➠
(5)d
dx
(
(3x)4x)
➠
(6)d
dxx (ex )
➠
(7)d
dxxπ+ πx
➠
(8)d
dx
(
1 +1
x
)x
➠
(9)d
dx(ln(x))x
➠
(10)d
dx(f (x)g(x)h(x)) ➠
Answers to Exercises
Answers for 1.1
1. (a) 8, (b) 6, (c) DNE, (d) −2, (e) −1, (f) 8, (g) 7, (h) 6, (i) 3, (j) −3/2, (k) 6, (l) 2 2.
1 3. 2 4. 3 5. 3/5 6. 0.6931 ≈ ln(2) 7. 2.718 ≈ e 8. Consider
what happens when x is near zero and positive, as compared to when x is near
zero and negative. 9. The limit does not exist, so it is not surprising that the
resulting values are so different. 10. When v approaches c from below, then tv
approaches zero—meaning that one second to the stationary observations seems
like very little time at all for our traveler.
Answers for 1.2
1. For these problems, there are many possible values of δ, so we provide an
inequality that δ must satisfy when ε = 0.1. (a) δ < 1/30, (b) δ <
√110
10− 1 ≈
0.0488, (c) δ < arcsin (1/10) ≈ 0.1002, (d) δ < arctan (1/10) ≈ 0.0997 (e) δ < 13/100,
(f) δ < 59/400 2. Let ε > 0. Set δ = ε. If 0 < |x − 0| < δ, then |x · 1| < ε, since
sin
(1
x
)
≤ 1, |x sin
(1
x
)
− 0| < ε. 3. Let ε > 0. Set δ = ε/2. If 0 < |x − 4| < δ,
then |2x − 8| < 2δ = ε, and then because |2x − 8| = |(2x − 5) − 3|, we conclude
|(2x − 5) − 3| < ε. 4. Let ε > 0. Set δ = ε/4. If 0 < |x − (−3)| < δ, then
| − 4x − 12| < 4δ = ε, and then because | − 4x − 12| = |(−4x − 11) − 1|, we conclude
|(−4x − 11) − 1| < ε. 5. Let ε > 0. No matter what I choose for δ, if x is within
δ of −2, then π is within ε of π. 6. As long as x , −2, we havex2 − 4
x + 2= x − 2,
and the limit is not sensitive to the value of the function at the point −2; the limit
calculus 97
only depends on nearby values, so we really want to compute limx→−2
(x − 2). Let ε > 0.
Set δ = ε. Then if 0 < |x − (−2)| < δ, we have |(x − 2) − (−4)| < ε. 7. Let ε > 0.
Pick δ so that δ < 1 and δ <ε
61. Suppose 0 < |x − 4| < δ. Then 4 − δ < x < 4 + δ.
Cube to get (4 − δ)3 < x3 < (4 + δ)3. Expanding the right-side inequality, we get
x3 < δ3+ 12 · δ2
+ 48 · δ + 64 < δ + 12δ + 48δ + 64 = 64 + ε. The other inequality is
similar. 8. Let ε > 0. Pick δ small enough so that δ < ε/6 and δ < 1. Assume
|x − 1| < δ, so 6 · |x − 1| < ε. Since x is within δ < 1 of 1, we know 0 < x < 2. So
|x + 4| < 6. Putting it together, |x + 4| · |x − 1| < ε, so |x2+ 3x − 4| < ε, and therefore
|(x2+ 3x − 1) − 3| < ε. 9. Let ε > 0. Set δ = 3ε. Assume 0 < |x − 9| < δ. Divide
both sides by 3 to get|x − 9|
3< ε. Note that
√x + 3 > 3, so
|x − 9|√
x + 3< ε. This
can be rearranged to conclude
∣∣∣∣∣∣
x − 9√
x − 3− 6
∣∣∣∣∣∣< ε. 10. Let ε > 0. Set δ to be the
minimum of 2ε and 1. Assume x is within δ of 2, so |x − 2| < 2ε and 1 < x < 3.
So
∣∣∣∣∣
x − 2
2
∣∣∣∣∣
< ε. Since 1 < x < 3, we also have 2x > 2, so
∣∣∣∣∣
x − 2
2x
∣∣∣∣∣
< ε. Simplifying,∣∣∣∣∣
1
2−
1
x
∣∣∣∣∣< ε, which is what we wanted.
Answers for 1.3
1. 7 2. 5 3. 0 4. DNE 5. 1/6 6. 0 7. 3 8. 172 9. 0
10. 2 11. DNE 12.√
2 13. 3a2 14. 512 15. −4
Answers for 1.4
1. −∞ 2. 3/14 3. 1/2 4. −∞ 5. ∞ 6. ∞ 7. 0 8. −∞ 9.
x = 1 and x = −3 10. x = −4
Answers for 1.5
1. 0 2. −1 3.1
24. −∞ 5. π 6. 0 7. 0 8. 17 9. After 10
years, ≈ 174 cats; after 50 years, ≈ 199 cats; after 100 years, ≈ 200 cats; after 1000
years, ≈ 200 cats; in the sense that the population of cats cannot grow indefinitely
this is somewhat realistic. 10. The amplitude goes to zero.
98
Answers for 1.6
1. f (x) is continuous at x = 4 but it is not continuous on R. 2. f (x) is continuous
at x = 3 but it is not continuous on R. 3. f (x) is not continuous at x = 1 and it is
not continuous on R. 4. f (x) is not continuous at x = 5 and it is not continuous
on R. 5. f (x) is continuous at x = −5 and it is also continuous on R. 6. R
7. (−∞,−4) ∪ (−4,∞) 8. (−∞,−3) ∪ (−3, 3) ∪ (3,∞) 9. x = −0.48, x = 1.31,
or x = 3.17 10. x = 0.20, or x = 1.35
Answers for 2.1
1. f (2) = 10 and f ′(2) = 7 2. p′(x) = s(x) and r′(x) = q(x) 3. f ′(3) ≈ 4 4.
f ′(−2) = 4 5. f (1.2) ≈ 2.2 6. (a) [0, 4.5) ∪ (4.5, 6], (b) [0, 3) ∪ (3, 6], (c) See
Figure 7. f ′(−3) = −6 with tangent line y = −6x − 13 8. f ′(1) = −1/9 with1 2 3 4 5 6
−1
1
x
y
Answer 2.1.6: (c) a sketch of f ′(x).
tangent line y =−1
9x +
4
99. f ′(5) =
1
2√
2with tangent line y =
1
2√
2x − 1
2√
2
10. f ′(4) =−1
16with tangent line y =
−1
16x +
3
4
Answers for 2.2
1. 0 2. 0 3. 0 4. 0 5. 100x99 6. −100x−101 7. −5x−6 8.
πxπ−1 9. (3/4)x−1/4 10. −(9/7)x−16/7 11. 15x2+ 24x 12. −20x4
+
6x + 10/x3 13. −30x + 25 14.3
2x−1/2 − x−2 − exe−1 15. −5x−6 − x−3/2/2
16. 3x2+ 6x − 1 17. 2x − 1 18. x−1/2/2 19. 4x3 − 4x 20. −49t/5+ 5,
−49/5 21. See Figure 22. x3/16 − 3x/4 + 4 23. y = 13x/4 + 5 24.
−2 −1 1 2
−10
−5
5
10
f (x)
cf (x)
f ′(x)
(cf (x))′
x
y
Answer 2.2.21.
y = 24x − 48 − π3 25.d
dxcf (x) = lim
h→0
cf (x + h) − cf (x)
h= c lim
h→0
f (x + h) − f (x)
h=
cf ′(x).
Answers for 3.1
1. min at x = 1/2 2. min at x = −1, max at x = 1 3. max at x = 2, min at
x = 4 4. min at x = ±1, max at x = 0. 5. min at x = 1 6. none 7. min
at x = 0, max at x =3 ±√
17
28. none 9. local max at x = 5 10. local
min at x = 49 11. local min at x = 0 12. one 13. if c ≥ 0, then there are
calculus 99
no local extrema; if c < 0 then there is a local max at x = −√
|c|3
and a local min at
x =
√
|c|3
Answers for 3.2
1. min at x = 1/2 2. min at x = −1, max at x = 1 3. max at x = 2, min
at x = 4 4. min at x = ±1, max at x = 0. 5. min at x = 1 6. none 7.
max at x = 0, min at x = ±11 8. f ′(x) = 2ax + b, this has only one root and
hence one critical point; a < 0 to guarantee a maximum.
Answers for 3.3
1. concave up everywhere 2. concave up when x < 0, concave down when x > 0
3. concave down when x < 3, concave up when x > 3 4. concave up when
x < −1/√
3 or x > 1/√
3, concave down when −1/√
3 < x < 1/√
3 5. concave
up when x < 0 or x > 2/3, concave down when 0 < x < 2/3 6. concave up
when x < 0, concave down when x > 0 7. concave up when x < −1 or x > 1,
concave down when −1 < x < 0 or 0 < x < 1 8. concave up on (0,∞) 9.
concave up on (0,∞) 10. concave up on (−∞,−1) and (0,∞) 11. up/incr:
(3,∞), up/decr: (−∞, 0), (2, 3), down/decr: (0, 2)
Answers for 3.4
1. min at x = 1/2 2. min at x = −1, max at x = 1 3. max at x = 2, min
at x = 4 4. min at x = ±1, max at x = 0. 5. min at x = 1 6. none 7.
none 8. max at −5−1/4, min at 5−1/4 9. max at −1, min at 1 10. min at
2−1/3
Answers for 3.5
1. y-intercept at (0, 0); no vertical asymptotes; critical points: x = ± 4√5; local
max at x = − 4√5, local min at x = − 4√
5; increasing on (−∞,− 4√5), decreasing on
(− 4√5,
4√5), increasing on (
4√5,∞); concave down on (−∞, 0), concave up on (0,∞);
root at x = 0; no horizontal asymptotes; interval for sketch: [−1.2, 1.2] (answers
100
may vary) 2. y-intercept at (0, 0); no vertical asymptotes; no critical points;
no local extrema; increasing on (−∞,∞); concave down on (−∞, 0), concave up
on (0,∞); roots at x = 0; no horizontal asymptotes; interval for sketch: [−3, 3]
(answers may vary) 3. y-intercept at (0, 0); no vertical asymptotes; critical points:
x = 1; local max at x = 1; increasing on [0, 1), decreasing on (1,∞); concave down
on [0,∞); roots at x = 0, x = 4; no horizontal asymptotes; interval for sketch:
[0, 6] (answers may vary) 4. y-intercept at (0, 0); no vertical asymptotes; critical
points: x = −3, x = −1; local max at x = −3, local min at x = −1; increasing on
(−∞,−3), decreasing on (−3,−1), increasing on (−1,∞); concave down on (−∞,−2),
concave up on (−2,∞); roots at x = −3, x = 0; no horizontal asymptotes; interval
for sketch: [−5, 3] (answers may vary) 5. y-intercept at (0, 5); no vertical
asymptotes; critical points: x = −1, x = 3; local max at x = −1, local min at
x = 3; increasing on (−∞,−1), decreasing on (−1, 3), increasing on (3,∞); concave
down on (−∞, 1), concave up on (1,∞); roots are too difficult to be determined—
cubic formula could be used; no horizontal asymptotes; interval for sketch: [−2, 5]
(answers may vary) 6. y-intercept at (0, 0); no vertical asymptotes; critical points:
x = 0, x =10 ±
√85
5; local max at x =
10 −√
85
5, local min at x =
10 +√
85
5;
increasing on (−∞,10 −
√85
5), decreasing on (
10 −√
85
5,10 +
√85
5), increasing on
(10 +
√85
5,∞); concave down on (−∞, 0), concave up on (0,
15 −√
195
10), concave
down on (15 −
√195
10,15 +
√195
10), concave up on (
15 +√
195
10,∞); roots at x = 0,
x =5 ±√
21
2; no horizontal asymptotes; interval for sketch: [−1, 5] (answers may
vary) 7. no y-intercept; vertical asymptote at x = 0; critical points: x = 0,
x = ±1; local max at x = −1, local min at 1; increasing on (−∞,−1), decreasing on
(−1, 0)∪ (0, 1), increasing on (1,∞); concave down on (−∞, 0), concave up on (0,∞);
no roots; no horizontal asymptotes; interval for sketch: [−2, 2] (answers may vary)
8. no y-intercept; vertical asymptote at x = 0; critical points: x = 0, x =13√
2;
local min at x =13√
2; decreasing on (−∞, 0), decreasing on (0,
13√
2), increasing on
(13√
2,∞); concave up on (−∞,−1), concave down on (−1, 0), concave up on (0,∞);
calculus 101
root at x = −1; no horizontal asymptotes; interval for sketch: [−3, 2] (answers may
vary)
Answers for 4.1
1. 3x2(x3 − 5x + 10) + x3(3x2 − 5) 2. (x2+ 5x − 3)(5x4 − 18x2
+ 6x − 7) + (2x +
5)(x5 − 6x3+ 3x2 − 7x + 1) 3. 2e2x 4. 3e3x 5. 6xe4x
+ 12x2e4x 6.−48ex
x17+
3ex
x167. f ′ = 4(2x − 3), y = 4x − 7 8. 3 9. 10 10. −13 11.
−5 12.d
dxf (x)g(x)h(x) =
d
dxf (x)(g(x)h(x)) = f (x)
d
dx(g(x)h(x))+ f ′(x)g(x)h(x) =
f (x)(g(x)h′(x)+g′(x)h(x))+f ′(x)g(x)h(x) = f (x)g(x)h′(x)+f (x)g′(x)h(x))+f ′(x)g(x)h(x)
Answers for 4.2
1.3x2
x3 − 5x + 10−
x3(3x2 − 5)
(x3 − 5x + 10)22.
2x + 5
x5 − 6x3 + 3x2 − 7x + 1−
(x2+ 5x − 3)(5x4 − 18x2
+ 6x − 7)
(x5 − 6x3 + 3x2 − 7x + 1)2
3.2xex − (ex − 4)2
4x24.
(x + 2)(−1 − (1/2)x−1/2) − (2 − x −√
x)
(x + 2)25. y = 17x/4−
41/4 6. y = 11x/16−15/16 7. y = 19/169−5x/338 8. −1/4 9. 8/9
10. 24 11. −3 12. f (4) = 1/3,d
dx
f (x)
g(x)= 13/18
Answers for 5.1
1. 4x3 − 9x2+ x + 7 2. 3x2 − 4x + 2/
√x 3. 6(x2
+ 1)2x 4.√
169 − x2 −x2/√
169 − x2 5. (2x − 4)√
25 − x2−(x2−4x+5)x/
√25 − x2 6. −x/
√r2 − x2 7. 2x3/
√1 + x4 8.
1
4√
x(5 −√
x)3/2
9. 6 + 18x 10.2x + 1
1 − x+
x2+ x + 1
(1 − x)211. −1/
√25 − x2 −
√25 − x2/x2 12.
1
2
(−169
x2− 1
) /√
169
x− x 13.
3x2 − 2x + 1/x2
2√
x3 − x2 − (1/x)14.
300x
(100 − x2)5/215.
1 + 3x2
3(x + x3)2/316.
4x(x2+ 1) +
4x3+ 4x
2√
1 + (x2 + 1)2
/
2
√
(x2 + 1)2 +
√
1 + (x2 + 1)2 17. 5(x+8)4 18. −3(4−x)2 19. 6x(x2+5)2
20. −12x(6 − 2x2)2 21. 24x2(1 − 4x3)−3 22. 5 + 5/x2 23. −8(4x −
102
1)(2x2 − x + 3)−3 24. 1/(x + 1)2 25. 3(8x − 2)/(4x2 − 2x + 1)2 26.
−3x2+ 5x − 1 27. 6x(2x − 4)3
+ 6(3x2+ 1)(2x − 4)2 28. −2/(x − 1)2
29. 4x/(x2+ 1)2 30. (x2 − 6x + 7)/(x − 3)2 31. −5/(3x − 4)2 32.
60x4+72x3
+18x2+18x−6 33. (5−4x)/((2x+1)2(x−3)2) 34. 1/(2(2+3x)2)
35. 56x6+ 72x5
+ 110x4+ 100x3
+ 60x2+ 28x + 6 36. y = 23x/96 − 29/96
37. y = 3 − 2x/3 38. y = 13x/2 − 23/2 39. y = 2x − 11 40. y =
20 + 2√
5
5
√
4 +√
5
x +3√
5
5
√
4 +√
5
Answers for 5.2
1. −x/y 2. x/y 3. −(2x + y)/(x + 2y) 4. (2xy− 3x2 − y2)/(2xy− 3y2 − x2)
5.2xy + x2
3y26. −
√y/√
x 7.y3/2 − 2
1 − y1/23x/28. −y2/x2 9. 1 10.
y = 2x ± 6 11. y = x/2 ± 3 12. (√
3, 2√
3), (−√
3,−2√
3), (2√
3,√
3),
(−2√
3,−√
3) 13. y = 7x/√
3−8/√
3 14. y = (−y1/31 x +y1/3
1 x1+x1/31 y1)/x1/3
1
15. (y − y1)/(x − x1) = (2x31 + 2x1y2
1 − x1)/(2y31 + 2y1x2
1 + y1)
Answers for 5.3
1. (x+1)3√
x4 + 5(3/(x+1)+2x3/(x4+5)) 2. (2x/x2
+5x)x2e5x 3. 2 ln(x)x ln(x)−1
4. (100 + 100 ln(x))x100x 5. (4 + 4 ln(3x))(3x)4x 6. ((ex )/x + ex ln(x))xex
7. πxπ−1+ πx ln(π) 8. (ln(1 + 1/x) − 1/(x + 1))(1 + 1/x)x 9. (1/ ln(x) +
ln(ln(x)))(ln(x))x 10. (f ′(x)/f (x) + g′(x)/g(x) + h′(x)/h(x))f (x)g(x)h(x)
calculus 103
Index
asymptote
horizontal, 27
vertical, 23
Binomial Theorem, 45
chain rule, 79
composition of functions, 14
concave down, 60
concave up, 60
concavity test, 61
constant rule, 44
continuous, 30
critical point, 54
derivative
limit definition, 36
notation, 37
of ex , 49
derivative rules
chain, 79
constant, 44
power, 45
product, 72
quotient, 75
sum, 47
e, 49
Euler’s number, 49
ex , 49
explicit function, 86
Fermat’s Theorem, 54
first derivative test, 57
horizontal asymptote, 27
implicit function, 86
implicit differentiation, 86
infinite limit, 23
inflection point, 61
Inverse Function Theorem, 89
limit
at infinity, 26
definition, 6
definition of the derivative, 36
infinite, 23
limit laws, 17
local maximum, 53
local minimum, 53
local extremum, 53
logarithmic differentiation, 91
one-sided limit, 8
product rule, 72
quotient rule, 75
second derivative test, 64
Squeeze Theorem, 19
sum rule, 47
tangent line
informal definition, 35
the derivative of the natural logarithm, 88
the power rule, 45
triangle inequality, 13
vertical asymptote, 23